Text Book Back Questions and Answers - Chapter 9 - Electro Chemistry - 12th Chemistry Guide Samacheer Kalvi Solutions

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Electro Chemistry
Textual EvaluationSolved
Multiple Choice Questions
Question 1.
The number of electrons that have a total charge of 9650 coulombs is
(a) $6.22 \times 10^{23}$
(b) $6.022 \times 10^{24}$
(c) $6.022 \times 10^{22}$
(d) $6.022 \times 10^{-34}$
Answer:
(c) $6.022 \times 10^{22}$
Hint: $\mathrm{IF}=96500 \mathrm{C}=1$ mole of $\mathrm{e}^{-}=6.023 \times 10^{23} \mathrm{e}^{-}$
$
9650 \mathrm{C}=\frac{6.22 \times 10^{23}}{96500} \times 9650=6.022 \times 10^{22}
$
Question 2.
Consider the following half cell reactions:
$
\begin{aligned}
& \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{\circ}=-1.18 \mathrm{~V} \\
& \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+}+\mathrm{e}^{-} \mathrm{E}=-1.51 \mathrm{~V}
\end{aligned}
$
The $\mathrm{E}$ for the reaction $3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}$, and the possibility of the forward reaction are respectively.
(a) $2.69 \mathrm{~V}$ and spontaneous
(b) -2.69 and non spontaneous
(c) $0.33 \mathrm{~V}$ and Spontaneous
(d) $4.18 \mathrm{~V}$ and non spontaneous
Answer:
(b) -2.69 and non spontaneous
Hint: $\mathrm{Mn}^{+}+2 \mathrm{e}^{-} \rightarrow \operatorname{Mn}\left(\mathrm{E}_{\text {red }}^0\right)=1.18 \mathrm{~V}$
$
\begin{aligned}
& 2\left[\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+}+\mathrm{e}^{-}\right]\left(\mathrm{E}_{\mathrm{ox}}^0\right)=-1.51 \mathrm{~V} \\
& 3 \mathrm{Mn}^{2+}+\rightarrow \mathrm{Mn}^{3+}+2 \mathrm{Mn}^{3+}+\left(\mathrm{E}_{\text {cell }}^0\right)=? \\
& \mathrm{E}_{\text {red }}^0=\left(\mathrm{E}_{\mathrm{ox}}^0\right)+\left(\mathrm{E}_{\text {cell }}^0\right) \\
& =-1.51-1.18 \text { and non spontaneous } \\
& =-2.69 \mathrm{~V}
\end{aligned}
$
Since $\mathrm{E}^{\circ}$ is - ve $\Delta \mathrm{G}$ is +ve and the given forward cell reaction is non - spontaneous.

Question 3.
The button cell used in watches function as follows
$\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Ag}_2 \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O}_{(1)} \rightleftharpoons 2 \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Zn}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{OH}^{-}$(aq) the half cell potentials are
$\mathrm{Ag}_2 \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O}_{(1)}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag}_{(\mathrm{S})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-} \mathrm{E}^{\circ}=034 \mathrm{~V}$. The cell potential will be
(a) $0.84 \mathrm{~V}$
(b) $1.34 \mathrm{~V}$
(c) $1.10 \mathrm{~V}$
(d) $0.42 \mathrm{~V}$
Answer:
(c) $1.10 \mathrm{~V}$
Hint: Anodic oxidation: (Reverse the given reaction)
$\left(\mathrm{E}_{\mathrm{ox}}^0\right)=0.76 \mathrm{~V}$ cathodic reduction
$
\mathrm{E}_{\text {cell }}^0=\left(\mathrm{E}_{\text {ox }}^0\right)+\left(\mathrm{E}_{\text {red }}^0\right)=0.76+0.34=1.1 \mathrm{~V}
$
Question 4.
The molar conductivity of a $0.5 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of $\mathrm{AgNO}_3$ with electrolytic conductivity of $5.76 \times 10^{-}$ ${ }^3 \mathrm{~S} \mathrm{~cm}^{-1}$ at $298 \mathrm{~K}$ is
(a) $2.88 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
(b) $11.52 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
(c) $0.086 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
(d) $28.8 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
Answer:
(b) $11.52 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
Hint:
$
\begin{aligned}
& A=\frac{k}{M} \times 10^{-3} \mathrm{~mol}^{-1} \mathrm{~m}^3 \\
& =\frac{5.76 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \times 10^{-3}}{0.5} \mathrm{~mol}^{-1} \mathrm{~m}^3=\frac{5.76 \times 10^{-3} \times 10^{-3} \times 10^6}{0.5} \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \mathrm{~cm}^3 \\
& =11.52 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$
Question 5.

Calculate $\mathrm{A}^0$ HOAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at $25^{\circ} \mathrm{C}$.
(a) 517.2
(b) 552.7
(c) 390.7
(d) 217.5
Answer:
(c) 390.7
Hint:
$
\begin{aligned}
& \left(\Lambda_{\infty}\right)_{\mathrm{HOAc}}=\left[\left(\Lambda^{\circ}\right)_{\mathrm{HCl}}+\left(\Lambda^{\circ}\right)_{\mathrm{NaOAc}}\right]-\left(\Lambda^{\circ}\right)_{\mathrm{NaCl}} \\
& (426.2+91)-(126.5)=390.7
\end{aligned}
$
Question 6.
Faradays constant is defined as
(a) charge carried by I electron
(b) charge carried by one mole of electrons
(c) charge required to deposit one mole of substance
(d) charge carried by $6.22 \times 10^{10}$ electrons
Answer:
(b) charge carried by one mole of electrons
Hint:
IF $=96500 \mathrm{C}=1$ charge of mole of $\mathrm{e}^{-}=$charge of $6.022 \times 10^{23} \mathrm{e}^{-}$

Question 7.
How many faradays of electricity are required for the following reaction to occur
$
\mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+}
$
(a) $5 \mathrm{~F}$
(b) $3 \mathrm{~F}$
(C) $\mathrm{IF}$
(d) $7 \mathrm{~F}$
Answer:
(a) $5 \mathrm{~F}$
Hint:
$
7 \mathrm{MnO}_4^{-}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}
$
5 moles of electrons i.e., $5 \mathrm{~F}$ charge is required.
Question 8 .
A current strength of $3.86 \mathrm{~A}$ was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of $\mathrm{Ca}$ is $40 \mathrm{~g} / \mathrm{mol}$ and $\mathrm{IF}=$ $96500 \mathrm{C})$
(a) 4
(b) 2
(c) 8
(d) 6
Answer:
(b) 2
Hint: $\mathrm{m}=$ ZIt
$41 \mathrm{~mm} \mathrm{40sec}=2500$ seconds
$
=\frac{40 x 3.86 x 2500}{2 x 96500}
$
$
\begin{aligned}
& \mathrm{Z}=\frac{m}{n x 96500}=\frac{40}{2 x 96500} \\
& =2 \mathrm{~g}
\end{aligned}
$

Question 9.
During electrolysis of molten sodium chloride, the time required to produce $0.1 \mathrm{~mol}$ of chlorine gas using a current of $3 \mathrm{~A}$ is
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Answer:
(b) 107.2 minutes
Hint: $\frac{m}{Z I}$ (mass of 1 mole of $\mathrm{Cl}_2$ gas $=71$ )
$\mathrm{t}=\frac{m}{Z I}$ mass of 0.1 mole of $\mathrm{Cl}_2$ gas $=7.1 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$
=\frac{7.1}{\frac{71}{2 \times 96500} \times 3}\left(2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2+2 \mathrm{e}^{-}\right)=\frac{2 \times 96500 \times 7.1}{71 \times 3}=6433.33 \mathrm{sec}=107.2 \mathrm{~min}
$

Question 10.
The number of electrons delivered at the cathode during electrolysis by a current of $1 \mathrm{~A}$ in 60 seconds is (charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )
(a) $6.22 \times 10^{23}$
(b) $6.022 \times 10^{20}$
(c) $3.75 \times 10^{20}$
(d) $7.48 \times 10^{23}$
Answer:
(c) $3.75 \times 10^{20}$
Hint: $\mathrm{Q}=\mathrm{It}$ $=1 \mathrm{~A} \times 60 \mathrm{~S}$
$96500 \mathrm{C}$ charge $6.022 \times 10^{23}$ electrons
$60 \mathrm{C}$ charge $=\frac{6.022 \times 10^{23}}{96500} \times 960$
$=3.744 \times 10^{20}$ electrons
Question 11.
Which of the following electrolytic solution has the least specific conductance?
(a) $2 \mathrm{~N}$
(b) $0.002 \mathrm{~N}$
(c) $0.02 \mathrm{~N}$
(d) $0.2 \mathrm{~N}$
Answer:
(b) $0.002 \mathrm{~N}$
Hint: In general, specific conductance of an electrolyte decreases with dilution. $\mathrm{SO}, 0.002 \mathrm{~N}$ solution has least specific conductance.
Question 12.
While charging lead storage battery
(a) $\mathrm{PbSO}_4$ on cathode is reduced to $\mathrm{Pb}$
(b) $\mathrm{PbSO}_4$ on anode is oxidised to $\mathrm{PbO}_4$
(c) $\mathrm{PbSO}_4$ on anode is reduced to $\mathrm{Pb}$
(d) $\mathrm{PbSO}_4$ on cathode is oxidised to $\mathrm{Pb}$
Answer:
(c) $\mathrm{PbSO}_4$ on anode is reduced to $\mathrm{Pb}$.
Hint: Charging: anode: $\mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$ (s) $+\mathrm{SO}_4^{-2}$ (aq)
Cathode: $\mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(1) \rightarrow \mathrm{PbO}_2(\mathrm{~s})+\mathrm{SO}_4^{-2}(\mathrm{aq})+2 \mathrm{e}_{-}$

Question 13.
Among the following cells
I. Leclanche cell
II. Nickel - Cadmium cell
III. Lead storage battery
IV. Mercury cell
Primary cells are
(a) I and IV
(b) I and III
(c) III and IV
(d) II and III
Answer:
(a) I and IV
Question 14.
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
(a) Zinc is lighter than iron
(b) Zinc has lower melting point than iron
(c) Zinc has lower negative electrode potential than iron
(d) Zinc has higher negative electrode potential than iron
Answer:
(d) Zinc has higher negative electrode potential than iron
Hint: $\mathrm{E}^0 \mathrm{Zn}^{+} \mid \mathrm{Zn}=-0.76 \mathrm{~V}$ and $\mathrm{E}^0 \mathrm{Fe}^{2+} \mid \mathrm{Fe}=0.44 \mathrm{~V}$. Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.
Question 15.
Assertion: pure iron when heated in dry air is converted with a layer Of rust.
Reason: Rust has the composition $\mathrm{Fe}_3 \mathrm{O}_4$
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(d) both assertion and reason are false.
Hint: Both are false
1. Dry air has no reaction with iron
2. Rust has the composition $\mathrm{Fe}_2 \mathrm{O}_3 \times \mathrm{H}_2 \mathrm{O}$
Question 16.
In $\mathrm{H}_2-\mathrm{O}_2$ fuel cell the reaction occur at cathode is

(a) $\mathrm{O}_2$ (g) $+2 \mathrm{H}_2 \mathrm{O}$ (l) $+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}$(aq)
(b) $\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_2 \mathrm{O}$ (1)
(c) $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
(d) $\mathrm{H}^{+}+\mathrm{e}^{-} \rightarrow 1 / 2 \mathrm{H}_2$
Answer:
(a) $\mathrm{O}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}$ (1) $+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(\mathrm{aq})$
Question 17.
The equivalent conductance of $\mathrm{M} / 36$ solution of a weak monobasic acid is $6 \mathrm{mho} \mathrm{cm}^2$ and at infinite dilution is $400 \mathrm{mho} \mathrm{cm}^2$. The dissociation constant of this acid is
(a) $1.25 \times 10^{-16}$
(b) $6.25 \times 10^{-6}$
(c) $1.25 \times 10^{-4}$
(d) $6.25 \times 10^{-5}$
Answer:
(b) $6.25 \times 10^{-6}$
Hint: $\alpha=\frac{6}{400}$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{C}=\frac{6}{400} \times \frac{6}{400} \times \frac{1}{36} \\
& =6.25 \times 10^{-6}
\end{aligned}
$
Question 18.
A conductivity cell has been calibrated with a $0.01 \mathrm{M}, 1: 1$ electrolytic solution (specific conductance $(\mathrm{K}=$ $1.25 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}$ ) in the cell and the measured resistance was $800 \Omega$ at $25^{\circ} \mathrm{C}$. The cell constant is,
(a) $10^{-1} \mathrm{~cm}^{-1}$
(b) $10^{-1} \mathrm{~cm}^{-1}$
(c) $1 \mathrm{~cm}^{-1}$
(d) $5.7 \times 10^{-12}$
Answer:
(c) $1 \mathrm{~cm}^{-1}$
Hint: $\mathrm{R}=\mathrm{p} \cdot \frac{1}{A}$
Cell constant $=\frac{R}{\rho}=\mathrm{k} . \mathrm{R}\left(\frac{1}{\rho}=k\right)=1.25 \times 10^{-3} \mathrm{f}^{-1} \mathrm{~cm}^{-1} \times 800 \Omega=1 \mathrm{~cm}^{-1}$

Question 19.
Conductivity of a saturated solution of a sparingly soluble salt $\mathrm{AB}$ (1:1 electrolyte) at $298 \mathrm{~K}$ is $1.85 \times 10^{-5}$ $\mathrm{S} \mathrm{m}^{-1}$. Solubility product of the saltAB at $298 \mathrm{~K}\left(\Lambda^0{ }_{\mathrm{m}}\right)_{\mathrm{AB}}=14 \times 10^{-3} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$.
(a) $5.7 \times 10^{-2}$
(b) $1.32 \times 10^{12}$
(c) $7.5 \times 10^{-12}$
(d) $1.74 \times 10^{-12}$
Answer:
(d) $1.74 \times 10^{-12}$

Question 20.
In the electrochemical cell: $\mathrm{Zn}\left|\mathrm{ZnSO}_4(0.01 \mathrm{M})\right|\left|\mathrm{CuSO}_4(1.0 \mathrm{M})\right| \mathrm{Cu}$, the emf of this Daniel cell is $\mathrm{E}_1$. When the concentration of $\mathrm{ZnSO}_4$ is changed to $1.0 \mathrm{M}$ and that $\mathrm{CuSO}_4$ changed to $0.01 \mathrm{M}$, the emf changes to $E_2$. From the followings, which one is the relationship between $E_1$ and $E_2$ ?
(a) $E_1 (b) $E_1>E_2$
(c) $E_2=0 \uparrow E_1$
(d) $E_1=E_2$
Answer:
(b) $E_1>E_2$
Hint:

$\begin{array}{ll}
\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]} & \\
\mathrm{E}_1=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{10^{-2}}{1} & \mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \\
\mathrm{E}_1=\mathrm{E}_{\text {cell }}^{\circ}+0.0591 \ldots \ldots \ldots \ldots . .(1) & \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \\
\mathrm{E}_2=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{1}{10^{-2}} & \mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}) \\
\mathrm{E}_2=\mathrm{E}_{\text {cell }}^{\circ}-0.0591 \ldots \ldots \ldots \ldots . .(2) & \\
\therefore \mathrm{E}_1>\mathrm{E}_2
\end{array}$

Question 21.
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:
$
\mathrm{BrO}_4^{-} \stackrel{1.82 \mathrm{~V}}{\longrightarrow} \mathrm{BrO}_3^{-} \stackrel{1.5 \mathrm{~V}}{\longrightarrow} \mathrm{HBrO} \stackrel{1.595 \mathrm{~V}}{\longrightarrow} \mathrm{Br}_2 \stackrel{1.0652 \mathrm{~V}}{\longrightarrow} \mathrm{Br}
$
Then the species undergoing disproportional is
(a) $\mathrm{Br}_2$
(b) $\mathrm{BrO}_4^{-}$
(c) $\mathrm{BrO}_3^{-}$
(d) $\mathrm{HBrO}$
Answer:
(d) $\mathrm{HBrO}$
Hint:

$
\begin{aligned}
& \left(\mathrm{E}_{\text {cell }}\right)_{\mathrm{A}}=-1.82+1.5=-0.32 \mathrm{~V} \\
& \left(\mathrm{E}_{\text {cell }}\right)_{\mathrm{B}}=-1.5+1.595=+0.095 \mathrm{~V} \\
& \left(\mathrm{E}_{\text {cell }}\right)_{\mathrm{C}}=1.595+1.0652=-0.529 \mathrm{~V}
\end{aligned}
$
The species undergoing disproportionation is $\mathrm{HBrO}$
Question 22.
For the cell reaction
$
2 \mathrm{Fe}^{3+}(\mathrm{aq})+{ }^{2 \mathrm{I}}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{I}_2(\mathrm{aq})
$
$\mathrm{EC}^0$ cell $=0.24 \mathrm{~V}$ at $298 \mathrm{~K}$. The standard Gibbs energy $\left(\Delta, \mathrm{G}^0\right)$ of the cell reactions is
(a) $-46.32 \mathrm{KJ} \mathrm{mol}^{-1}$
(b) $-23.16 \mathrm{KJ} \mathrm{mol}^{-1}$
(c) $46.32 \mathrm{KJ} \mathrm{mol}^{-1}$
(d) $23.16 \mathrm{KJ} \mathrm{mor}^{-1}$
Answer:
(a) $-46.32 \mathrm{KJ} \mathrm{mol}^{-1}$
Question 23.
A certain current liberated $0.504 \mathrm{gm}$ of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?
(a) 31.75
(b) 15.8
(c) 7.5
(d) 63.5
Answer:
(b) 15.8
Question 24.
A gas $\mathrm{X}$ at $1 \mathrm{~atm}$ is bubble through a solution containing a mixture of $1 \mathrm{MY}^{-}$and $1 \mathrm{MZ}^{-1}$ at $25^{\circ} \mathrm{C}$. If the reduction potential of $Z>Y>X$, then
(a) $\mathrm{Y}$ will oxidize $\mathrm{X}$ and not $\mathrm{Z}$
(b) $\mathrm{Y}$ will oxidize $\mathrm{Z}$ and not $\mathrm{X}$
(c) $\mathrm{Y}$ will oxidize both $\mathrm{X}$ and $\mathrm{Z}$
(d) $\mathrm{Y}$ will reduce both $\mathrm{X}$ and $\mathrm{Z}$
Answer:
(a) $\mathrm{Y}$ will oxidize $\mathrm{X}$ and not $\mathrm{Z}$
Question 25.
Cell equation: $\mathrm{A}^{2+}+2 \mathrm{~B}^{-} \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}$

$\mathrm{A}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{AE}^{\circ}=+0.34 \mathrm{~V}$ and $\log _{10} \mathrm{~K}=15.6$ at $300 \mathrm{~K}$ for cell reactions find $\mathrm{E}^{\circ}$ for $\mathrm{B}^1+\mathrm{e}^{-} \rightarrow \mathrm{B}$
(a) 0.80
(b) 1.26
(c) -0.54
(d) -10.94
Answer:
(a) 0.80

Short Answer
Question 1.
Define anode and cathode
Answer:
1. Anode: The electrode at which the oxidation occur is called anode.
2. Cathode: The electrode at which the reduction occur is called cathode.
Question 2.
Why does conductivity of a solution decrease on dilution of the solution?
Answer:
Conductivity always decreases with decrease in concentration (on dilution of the solution) both for weak as much as for strong electrolytes. $j$ t is because the number of ions per unit volume that carry the current is a solution decreases on dilution.
Question 3.
State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.
Answer:
Kohlrausch's law:
It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Determination of the molar conductivity of weak electrolyte at infinite dilution. It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraushs Law. For example, the molar conductance of $\mathrm{CH}_3 \mathrm{COOH}$, can be calculated using the experimentally determined molar conductivities of strong electrolytes $\mathrm{HCI}, \mathrm{NaCI}$ and $\mathrm{CH}_3 \mathrm{COONa}$.

Question 4.
Describe the electrolysis of molten NaCI using inert electrodes.
Answer:
1. The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key.
2. The electrode which is attached to the negative end of the power supply is called the cathode and the one is which attached to the positive end is called the anode.
3. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time pull the electrons from the anode.
Cell reactions:
$\mathrm{Na}^{+}$ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.
Cathode (reduction)
$\mathrm{Na}_{(\mathrm{I})}^{+}+\mathrm{e}^{-} \mathrm{Na}(1)$
$\mathrm{E}^0=-2.71 \mathrm{~V}$
Similarly, $\mathrm{Cl}^{-}$ions are attracted towards anode where they losses their electrons and oxidised to chlorine gas. Anode (oxidation)
$2 \mathrm{Cl}^{-}(1) \mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-}$
$\mathrm{E}^{\circ}=-1.36 \mathrm{~V}$
The overall reaction is,
$
\begin{aligned}
& 2 \mathrm{Na}_{(1)}^{+}+2 \mathrm{Cl}_{(1)}^{-} \rightarrow 2 \mathrm{Na}_{(1)}+\mathrm{Cl}_{2(\mathrm{~g})}(\mathrm{g}) \\
& \mathrm{E}^0=4.07 \mathrm{~V}
\end{aligned}
$

The negative $\mathrm{E}^{\circ}$ value shows that the above reaction is a non spontaneous one. Hence, we have to supply a voltage greater than $4.07 \mathrm{~V}$ to cause the electrolysis of molten $\mathrm{NaCI}$. In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and anode is +ve.
Question 5.
State Faraday's Laws of electrolysis.
Answer:
Faraday's laws of electrolysis:
1. First law:
The mass of the substance (M) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell. $\mathrm{M} \alpha \mathrm{Q}$
2. Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents. $\mathrm{M} \alpha \mathrm{Z}$
Question 6.
Describe the construction of Daniel cell. Write the cell reaction.
Answer:
The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.
Oxidation half cell:
The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell:
A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.
Joining the half cell:
The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using
The ions of inert electrolyte do not react with other ions present in the half cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge Voltmeter
cannot get poured out, but through which the ions can move into (or) out of the half cells. When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.

Anodic oxidation:
The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at
zinc electrode, i.e., zinc is oxidised to $\mathrm{Zn}^2$ ions and the electrons. The $\mathrm{Zn}^2$ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip.
Electrons are liberated at zinc electrode and hence it is negative ( - ve).
$\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$(loss of electron-oxidation)
Cathodic reduction:
As discussed earlier, the electrons flow through the circuit from zinc to copper, where the $\mathrm{Cu}^{2+}$ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve).
$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})}$ (gain of electron-reduction)
Salt bridge:
The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in the increase in concentration of $\mathrm{Zn}^{2+}$ in solution. i.e., the solution contains more number of $\mathrm{Zn}^{2+}$ ions as compared to $\mathrm{SO}_4{ }^{2-}$ and hence the solution in the anodic compartment would become positively charged.
Similarly, the solution in the cathodic compartment would become negatively charged as the $\mathrm{Cu}^{2+}{ }^2$ ions are reduced to copper i.e., the cathodic solution contain more number of $\mathrm{SO}^{2-}{ }_4$ ions compared to $\mathrm{Cu}^{2+}$.
Completion of circuit:
Electrons flow from the negatively charged zinc anode into the positively charged copper electrode through the external wire, at the same time, anions move towards anode and cations are move towards the cathode compartment. This completes the circuit.
Consumption of Electrodes:
As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted in to $\mathrm{Zn}^{2+}$ the entire $\mathrm{Cu}^{2+}$ ions are converted in to metallic copper.
Daniel cell is represented as
$
\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|\left|\mathrm{Cu}^{2+}(\mathrm{aq})\right| \mathrm{Cu}_{(\mathrm{s})}
$

Question 7.
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Answer:
A galvanic cell works basically in reverse to an electrolytic cell. The anode is the electrode where oxidation takes place, in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.
The anode is also the electrode where metal atoms give up their electrons to the metal and go into solution. The electron left behind on it render $j$ t effectively negative and the electron flow goes from it through the wire to the cathode.
Positive aqueous ions in the solution are reduced by the incoming electrons on the cathode. This why the cathode is a positive electrode, because positive ions are reduced to metal atoms there.
Question 8.
The conductivity of a $0.01 \% \mathrm{M}$ solution of a $1: 1$ weak electrolyte at $298 \mathrm{~K}$ is $1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$.
1. molar conductivity of the solution
2. degree of dissociation and the dissociation constant of the weak electrolyte
Given that
$\lambda^0$ cation $=248.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\lambda^0$ anion $=51.8 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
Answer:
1. Molar conductivity
$\mathrm{C}=0.01 \mathrm{M}$
$\mathrm{k}=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$
(or)
$\mathrm{K}=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1}$
$\frac{\kappa \times 10^{-3}}{\mathrm{C}} \mathrm{S} \mathrm{m} \mathrm{mol}^{-1} \mathrm{~m}^3=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$
$\Lambda_{\mathrm{m}}=1.5 \times 10^{-3} \mathrm{~s} \mathrm{~m}^{-1}$
2. Degree of dissociation
$
\begin{aligned}
& \alpha=\frac{\Lambda_{\mathrm{m}}^{\circ}}{\Lambda_{\infty}^{\circ}}(\text { or }) \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \\
& =(248.2+51.8) \mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-1} \\
& =300 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\
& \mathrm{~K}_{\mathrm{a}}=\frac{\alpha^2 C}{1-\alpha} \\
& =\frac{(0.05)^2(0.01)}{1-0.05} \\
& \mathrm{~K}_{\mathrm{a}}=2.63 \times 10^{-5}
\end{aligned}
$

Question 9 .
Which of $0.1 \mathrm{M} \mathrm{HCI}$ and $0.1 \mathrm{M} \mathrm{KCI}$ do you expect to have greater molar conductance and why?
Answer:
Compare to $0.1 \mathrm{M} \mathrm{HCI}$ and $0.1 \mathrm{M} \mathrm{KCI}, 0.1 \mathrm{M} \mathrm{HCl}$ has greater molar conductance.
1. Molar conductance of $0.1 \mathrm{M} \mathrm{HCl}=39.132 \times 10^{-3} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$.
2. Molar conductance of $0.1 \mathrm{M} \mathrm{KCl}=12.896 \times 10^{-3} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$
Because, $\mathrm{H}^{+}$ion in aqueous solution being smaller size than $\mathrm{K}^{+}$ion and $\mathrm{H}^{+}$ion have greater mobility than $\mathrm{K}$ ion. When mobility of the ion increases, conductivity of that ions also increases. Hence, 0. 1M HCI solution has greater molar conductance than $0.1 \mathrm{M} \mathrm{KCI}$ solution.
Question 10.
Arrange the following solutions in the decreasing order of specific conductance.
1. $0.01 \mathrm{MKCI}$
2. $0.005 \mathrm{M} \mathrm{KCI}$
3. $0.1 \mathrm{M} \mathrm{KCI}$
4. $0.25 \mathrm{MKCI}$
5. $0.5 \mathrm{M} \mathrm{KCI}$
Answer:
$0.005 \mathrm{M} \mathrm{KCl}>0.01 \mathrm{M} \mathrm{KCI}>0.1 \mathrm{M} \mathrm{KCI}>0.25 \mathrm{KCl}>0.5 \mathrm{KCI}$.
Specific conductance and concentration of the electrolyte. So if concentration decreases, specific conductance increases.
Question 11.
Why is AC current used instead of DC in measuring the electrolytic conductance?
Answer:
1. AC current to prevent electrolysis of the solution.
2. If we apply DC current to the cell the positive ions will be attracted to the negative plate and the negative ions to the positive plate. This will cause the composition of the electrolyte to change while measuring the equivalent conductance.
3. So DC current through the conductivity cell will lead to the electrolysis of the solution taken in the cell. To avoid such a electrolysis, we have to use AC current for measuring equivalent conductance.

Question 12 .
$0.1 \mathrm{M} \mathrm{NaCI}$ solution is placed in two different cells having cell constant 0.5 and $0.25 \mathrm{~cm}^{-1}$ respectively. Which of the two will have greater value of specific conductance.
Answer:
The specific conductance values are same. Because the reaction (cation) of cell constant does not change.
Question 13 .
A current of $1.608 \mathrm{~A}$ is passed through $250 \mathrm{~mL}$ of $0.5 \mathrm{M}$ solution of copper sulphate for 50 minutes.
Calculate the strength of $\mathrm{Cu}^{2+}$ after electrolysis assuming volume to be constant and the current efficiency is $100 \%$.
Answer:
Given, I = I .608A
$\mathrm{t}=50 \min$ (or) $50 \times 60=3000 \mathrm{~S}$
$\mathrm{V}=250 \mathrm{~mL}$
$\mathrm{C}=0.5 \mathrm{M}$
$\eta=100 \%$
The number of Faraday's of electricity passed through the $\mathrm{CuSO}_4$ solution
$
\begin{aligned}
& Q=\text { It } \\
& =Q=1.608 \times 3000 \\
& Q=4824 \mathrm{C}
\end{aligned}
$
Number of Faraday's of electricity $=\frac{4824 \mathrm{C}}{96500 \mathrm{C}}=0.5 \mathrm{~F}$
Electrolysis of $\mathrm{CuSO}_4$
$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})}$
The above equation shows that $2 \mathrm{~F}$ electricity will deposit 1 mole of $\mathrm{Cu}^{2+}$
$0.5 \mathrm{~F}$ electicity will deposit $\frac{1 \mathrm{~mol}}{2 F} \times 0.5 \mathrm{~F}=0.025 \mathrm{~mol}$
Initial number of molar of $\mathrm{Cu}^{2+}$ in $250 \mathrm{ml}$ of solution $=\frac{1 \mathrm{~mol}}{250 \mathrm{~mL}} \times 250 \mathrm{~mL}=0.125 \mathrm{~mol}$
Number of molar of $\mathrm{Cu}^{2+}$ after electrolysis $0.125-0.025=0.1 \mathrm{~mol}$
Concentration of $\mathrm{Cu}^{2+}=\frac{0.1 \mathrm{~mol}}{250 \mathrm{~mL}} \times 1000 \mathrm{~mL}=0.4 \mathrm{M}$

Question 14.
Can $\mathrm{Fe}^{3+}$ oxidises Bromide to bromine under standard conditions?
Given:

$\mathbf{E}_{\mathrm{Fe}^{3+}}^{\circ} \mid \mathrm{Fe}^{2+}=0.771 \quad \mathrm{E}_{\mathrm{Br}_2 \mid \mathrm{Br}^*}^{\circ}=-1.09 \mathrm{~V}$
Answer:
Required half cell reaction
$
\begin{array}{ll}
2 \mathrm{Br}^{-} \rightarrow \mathrm{Br}^2+2 \mathrm{e}^{-} & \left(\mathrm{E}_{\mathrm{ox}}^{\circ}\right)=-1.09 \mathrm{~V} \\
2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Fe}^{2+} & \left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)=+0.771 \mathrm{~V} \\
2 \mathrm{Fe}^{3+}+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Br}_2 & \left(\mathrm{E}_{\mathrm{cell}}^{\circ}\right)=? \\
\mathrm{E}_{\text {cell }}^0=\left(\mathrm{E}_{\mathrm{ox}}^0\right)+\left(\mathrm{E}_{\text {red }}^0\right)=-1.09+0.771=-0.319 \mathrm{~V}
\end{array}
$
We know that $\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^0$ cell
If $\mathrm{E}_{\text {cell }}^0$ is -ve; $\Delta \mathrm{G}$ is +ve and the cell reaction is non-spontaneous.
Hence, $\mathrm{Fe}^{3+}$ cannot oxidise Bromide to Bromine.

Question 15.
Is it possible to store copper sulphate in an iron vessel for a long time?
Given:
$
\left.\mathrm{E}^{\circ} \mathrm{Cu}^{2+}\right|_{\mathrm{Cu}}=0.34 \mathrm{~V} \text { and }\left.\mathrm{E}^{\circ} \mathrm{Fe}^{2+}\right|_{\mathrm{Fe}}=+0.44 \mathrm{~V}
$
Answer:
$
\mathrm{E}_{\text {cell }}^0=\left(\mathrm{E}_{\text {ox }}^0\right)+\left(\mathrm{E}_{\text {red }}^0\right)=0.44 \mathrm{~V}+0.34 \mathrm{~V}=0.78 \mathrm{~V}
$
These $+v e E_{\text {cell }}^0$ values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 16.
Two metals $\mathrm{M}_1$ and $\mathrm{M}_2$ have reduction potential values of $-\mathrm{xV}$ and $+\mathrm{yV}$ respectively. Which will liberate $\mathrm{H}_2$ in $\mathrm{H}_2 \mathrm{SO}_4$ ?
Answer:
Metals having negative reduction potential acts as powerful reducing agent. Since $\mathrm{M}_1$ has $-\mathrm{xV}$, therefore M1 easily liberate $\mathrm{H}_2$ in $\mathrm{H}_2 \mathrm{SO}_4$. Metals having higher oxidation potential will liberate $\mathrm{H}_2$ from $\mathrm{H}_2 \mathrm{SO}_4$. Hence, the metal $\mathrm{M}_1$ having $+\mathrm{xV}$, oxidation potential will liberate $\mathrm{H}_2$ from $\mathrm{H}_2 \mathrm{SO}_4$.

Question 17.
Reduction potential of two metals $\mathrm{M}_1$ and $\mathrm{M}_2$ are
$
\mathbf{E}^{\circ} \mathrm{M}_1^{2+} \mathrm{M}_1=-2.3 \mathrm{~V} \text { and } \mathrm{E}^{\circ} \mathrm{M}_2^{2+} \mathrm{M}_2=0.2 \mathrm{~V}
$
Predict which one Is better for coating the surface of iron.
Given:

$
\left.\mathrm{E}^{\circ} \mathrm{Fe}^{2+}\right|_{\mathrm{Fe}}=-0.44 \mathrm{~V}
$
Answer:
Oxidation potential of $\mathrm{M}_1$ is more +ve than the oxidation potential of $\mathrm{Fe}$ which indicates that it will prevent iron from rusting.

Question 18 .
Calculate the standard emf of the cell: $\mathrm{Cd}\left|\mathrm{Cd}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}$ and determine the cell reaction. The standard reduction potentials of $\mathrm{Cu}^{2+} \mid \mathrm{Cu}$ and $\mathrm{Cd}^{2+} \mid \mathrm{Cd}$ are $0.34 \mathrm{~V}$ and -0.40 volts respectively. Predict the feasibility of the cell reaction.
Answer:
Cell reactions:

$\mathrm{Cd}(\mathrm{s})+2 \mathrm{e}^{-} \mathrm{Cd}^{2+}+\mathrm{Cu}(\mathrm{s})$
$\mathrm{E}_{\text {cell }}^0=\left(\mathrm{E}_{\text {ox }}^0\right)+\left(\mathrm{E}_{\text {red }}^0\right)=0.4+0.34$
emf is $+v e$, so $\Delta \mathrm{G}$ is (-)ve, the reaction is feasible.
Question 19.
In fuel cell $\mathrm{H}_2$ and $\mathrm{O}_2$ react to produce electricity. In the process, $\mathrm{H}_2$ gas is oxidised at the anode and $\mathrm{O}_2$ at cathode. If 44.8 litre of $\mathrm{H}_2$ at $25^{\circ} \mathrm{C}$ and also pressure reacts in 10 minutes, what is average current
produced? If the entire current is used for electro deposition of $\mathrm{Cu}$ from $\mathrm{Cu}^{2+}$, how many grams of $\mathrm{Cu}$ deposited?
Answer:
Oxidation at anode:
$
2 \mathrm{H}_2(\mathrm{~g})+4 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 4 \mathrm{H}_2 \mathrm{O}(1)+4 \mathrm{e}^{-}
$
1 mole of hydrogen gas produces 2 moles of electrons at $25^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure, 1 mole of hydrogen gas occupies $=22.4$ litres
$\therefore$ no. of moles of hydrogen gas produced $=\frac{1 \text { mole }}{22.4 \text { litres }} \times 44.8$ litres $=2$ moles of hydrogen
$\therefore 2$ of moles of hydrogen produces 4 moles of electro i.e., $4 \mathrm{~F}$ charge. We know that $Q=$ It
$\mathrm{I}=\frac{Q}{t}=\frac{4 \mathrm{~F}}{10 \mathrm{mins}}=\frac{4 \times 96500}{10 x 60 s}$
$\mathrm{I}=643.33 \mathrm{~A}$
Electro deposition of copper
$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})}$
$2 \mathrm{~F}$ charge is required to deposit
1 mole of copper i.e., $63.5 \mathrm{~g}$
If the entire current produced in the fuel cell i.e., $4 \mathrm{~F}$ is utilised for electrolysis, then $2 \times 63.5$ i.e., $127.0 \mathrm{~g}$ copper will be deposited at cathode.

Question 20 .
The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If $2.935 \mathrm{~g}$ of $\mathrm{Ni}$ was deposited in the first cell. The amount of $\mathrm{Cr}$ deposited in the another cell? Given: molar mass of Nickel and chromium are 58.74 and $52 \mathrm{gm}^{-1}$ respectively.
Answer:
$
\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{s})
$
$
\mathrm{Cr}^{2+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s})
$
The above reaction indicates that $2 \mathrm{~F}$ charge is required to deposit $58.7 \mathrm{~g}$ of Nickel form nickel nitrate and $3 \mathrm{~F}$ charge is required to deposit $52 \mathrm{~g}$ of chromium. Given that 2.935 gram of Nickel is deposited $2 \mathrm{~F}$ The amount of charge passed through the cell $=\frac{2 F}{58.7 g} \times 2.935 \mathrm{~g}=0.1 \mathrm{~F}$
If 0 . IF charge is passed through chromium nitrate the amount of chromium deposited $=52 \mathrm{~g} \mathrm{x} 0 . \mathrm{IF}=1.733 \mathrm{~g}$
Question 21.
$0.1 \mathrm{M}$ copper sulphate solution in which copper electrode is dipped at $25 \mathrm{C}$. Calculate the electrode potential of copper.
Answer:
Given that
$
\begin{aligned}
& {\left[\mathrm{Cu}^{2+}\right]=0.1 \mathrm{M}} \\
& \mathrm{E}^0 \mathrm{Cu}^{2+} \mid \mathrm{Cu}=0.34 \\
& \mathrm{E}_{\mathrm{cell}}=?
\end{aligned}
$
Cell reaction is $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$
$
\begin{aligned}
& \mathrm{E}_{\text {cell }}=\mathrm{E}^0-\frac{0.0591}{n} \log \frac{[\mathrm{Cu}]}{\left[\mathrm{Cu}^{2+}\right]}=0.34-\frac{0.0591}{2} \log \frac{1}{0.1} \\
& =0.34-0.0296=0.31 \mathrm{~V}
\end{aligned}
$

Question 22.
For the cell $\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}) \| \mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}$ (s), calculate the equilibrium constant at $25^{\circ} \mathrm{C}$ and maximum work that can be obtained during operation of cell. Given:
$
\left.\mathrm{E}^0 \mathrm{Mg}^{2+}\right|_{\mathrm{Mg}}=+2.37 \mathrm{~V} \text { and } \mathrm{E}_{\left.\mathrm{Ag}^0\right|_{\mathrm{Ag}}}=0.80 \mathrm{~V}
$
Answer:
Oxidation at anode $\quad \mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \quad\left(\mathrm{E}_{\mathrm{ox}}^{\circ}\right)=2.37 \mathrm{~V}$
Reduction at cathode $\quad \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \quad\left(\mathrm{E}_{\text {red }}^{\circ}\right)=0.80 \mathrm{~V}$
$
\mathrm{E}_{\text {cell }}^0=\left(\mathrm{E}_{\text {ox }}^0\right)+\left(\mathrm{E}_{\text {red }}^0\right)=2.37+0.80=3.17 \mathrm{~V}
$
Overall reaction
$
\begin{aligned}
& \mathrm{Mg}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{Ag} \\
& \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\
& =-2 \times 96500 \times 3.17 \\
& =-6.118 \times 10^5 \mathrm{~J}
\end{aligned}
$
We know that $\mathrm{W}_{\max }=\Delta \mathrm{G}^{\circ}$

$
\mathrm{W}_{\max }=+6.118 \times 10_5 \mathrm{~J}
$
Relationship between $\Delta \mathrm{G}^{\circ}$ and $\mathrm{K}_{\mathrm{eq}}$ is,
$
\begin{aligned}
& \Delta \mathrm{G}=-2.303 \mathrm{RT} \log \mathrm{K}_{\text {eq }} \\
& \Delta \mathrm{G}=-2.303 \times 8.314 \times 298 \log \mathrm{K}_{\mathrm{eq}}\left[25^{\circ} \mathrm{C}=298 \mathrm{~K}\right] \\
& \log \mathrm{K}_{\mathrm{eq}}=\frac{6.118 \times 10^5}{2.303 \times 8.314 \times 298}=\frac{6.118 \times 10^5}{5705.84} \\
& \log \mathrm{K}_{\mathrm{eq}}=107.223 \\
& \mathrm{~K}_{\mathrm{eq}}=\text { Antilog }(107.223)
\end{aligned}
$

Question 23.
$8.2 \times 10^{12}$ litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of $2 \times 10^6 \mathrm{Cs}^{-1}$ at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.
Answer:
Hydrolysis of water:
At anode: $2 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{H}^{+}+\mathrm{O}_2+4 \mathrm{e}^{-}$
At cathode: $2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$
Overall reaction: $6 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{H}^{-}+4 \mathrm{OH}^{-}+2 \mathrm{H}_2+\mathrm{O}_2$
(or)
Equation (1) $+(2) \times 2$
$=2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_2+\mathrm{O}_2$
According to Faraday's Law of electrolysis, to electrolyse two mole of Water $\left(36 \mathrm{~g} \simeq 36 \mathrm{~mL}\right.$. of $\left.\mathrm{H}_2 \mathrm{O}\right)$, 4F charge is required alternatively, when $36 \mathrm{~mL}$ of water is electrolysed, the charge generated $=4 \times 96500 \mathrm{C}$.
When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to
$
\frac{4 \times 96500 \mathrm{C}}{36 \mathrm{~mL}} \times 9 \times 10^{12} \mathrm{~L}=\frac{4 \times 96500 \times 9 \times 10^{12}}{36 \times 10^{-3}} \mathrm{C}=96500 \times 10^{15} \mathrm{C}
$
Given that in 1 second, $2 \times 10^6 \mathrm{C}$ is generated therefore, the time required to generate
$
\begin{aligned}
& 96500 \times 10^{15} \mathrm{C} \text { is }=\frac{1 \mathrm{~S}}{2 \times 10^6 \mathrm{C}} \times 96500 \times 10^{15} \mathrm{C}=48250 \times 10^9 \mathrm{~S} \\
& \text { Number of year }=\frac{48250 \times 10^9}{365 \times 24 \times 60 \times 60} 1 \text { year }=365 \text { days } \\
& =1.5299 \times 10^6 \\
& =365 \times 24 \text { hours } \\
& =365 \times 24 \times 60 \mathrm{~min} \\
& =365 \times 24 \times 60 \times 60 \mathrm{sec}
\end{aligned}
$
Question 24.
Derive an expression for Nernst equation.
Answer:
Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction.
Let us consider an electrochemical cell for which the overall redox reaction is,
$
\mathrm{xA}+\mathrm{yB}=1 \mathrm{C}+\mathrm{mD}
$

The reaction quotient $\mathrm{Q}$ is,
$
\frac{[\mathrm{C}]^1[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^y}
$
We know that,
$
\begin{aligned}
& \Delta \mathrm{G}=\Delta \mathrm{G}^0+\mathrm{RT} \ln \mathrm{Q} \\
& \Delta \mathrm{G}=-\mathrm{nFE}_{\text {cell }} \\
& \Delta \mathrm{G}^0=-\mathrm{nFE}_{\text {cell }}^0
\end{aligned}
$
equation (1) becomes
$
-\mathrm{nFE}_{\text {cell }}=-\mathrm{nFE}_{\text {cell }}^0+\mathrm{RT} \ln \mathrm{Q}
$
Subsitute the $\mathrm{Q}$ value in equation (2)
$
-\mathrm{nFE}_{\text {cell }}=-\mathrm{nFE}_{\text {cell }}^0+\mathrm{RT} \ln \frac{[\mathrm{C}]^1[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}
$
Divide the whole equation (3) by $-\mathrm{nF}$
$
\begin{aligned}
& \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ} \text { cell }-\frac{R T}{n F} \ln \left(\frac{[\mathrm{C}]^{\mathrm{1}}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right) \\
& \mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ} \text { cell }-\frac{2.303 R T}{n F} \log \left(\frac{[\mathrm{C}]^{\mathrm{1}}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)
\end{aligned}
$
This is called the Nernst equation.
At $25^{\circ} \mathrm{C}(298 \mathrm{~K})$ equation (4) becomes,
$
\begin{aligned}
& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{2.303 \times 8.314 \times 298}{n x 96500} \log \left(\frac{[\mathrm{C}]^1[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right) \\
& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \left(\frac{[\mathrm{C}]^1[\mathrm{D}]^{\mathrm{mI}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)
\end{aligned}
$

Question 25.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection. $\mathrm{Al}, \mathrm{Zn}$ and $\mathrm{Mg}$ are used as sacrificial anodes.
Question 26.
Explain the function of $\mathrm{H}_2-\mathrm{O}_2$ fuel cell.
Answer:
In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous $\mathrm{KOH}$ maintained at $200^{\circ} \mathrm{C}$ and $20-40 \mathrm{~atm}$. Porous graphite electrode containing $\mathrm{Ni}$ and $\mathrm{NiO}$ serves as the inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.
Oxidation occurs at the anode:
$
2 \mathrm{H}_{2(\mathrm{~g})}+4 \mathrm{OH}_{-(\mathrm{aq})} \rightarrow 4 \mathrm{H}_2 \mathrm{O}_{(1)}+4 \mathrm{e}^{-}
$
Reduction occurs at the cathode $\mathrm{O}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(1)+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}$(aq)
The overall reaction is $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(1)$
The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like $\mathrm{H}_2-\mathrm{O}_2$ fuel cell. Other fuel cell like propane $-\mathrm{O}_2$ and methane $\mathrm{O}_2$ have also been developed.

Question 27.
Ionic conductance at infinite dilution of $\mathrm{Al}^{3+}$ and $\mathrm{SO}_4{ }^{2-}$ are 189 and $160 \mathrm{mho} \mathrm{cm}^2$ equiv ${ }^{-1}$. Calculate the equivalent and molar conductance of the electrolyte $\mathrm{Al}_2\left(\mathrm{SO}_4\right)$ at infinite dilution.
Answer:
1. Molar conductance
$
\begin{aligned}
& \left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3}=2\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Al}^{3+}}+3\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{SO}_4^{2-}} \\
& =(2 \times 189)+(3 \times 160) \\
& =378+480 \\
& =858 \mathrm{mho} \mathrm{cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$
2. Equivalent conductnace
$
\begin{aligned}
& \left(\lambda_{\infty}\right)_{\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3}=\frac{1}{3}\left(\lambda_{\infty}\right)_{\mathrm{Al}^{3+}}+\frac{1}{2}\left(\lambda_{\infty}\right)_{\mathrm{SO}_4^{2-}} \\
& =\frac{189}{3}+\frac{160}{2} \\
& \left.=143 \mathrm{mho} \mathrm{cm}^2 \text { (g equiv }\right)^{-1}
\end{aligned}
$

$\begin{aligned}
& \Lambda=\frac{\kappa_2\left(\mathrm{Sm}^{-1}\right) \times 10^{-3}(\text { gram equivalent })^{-1} \mathrm{~m}^3}{\mathrm{~N}}=\frac{0.25 \times 10^{-3} \mathrm{~S}(\text { gram equivalent })^{-1} \mathrm{~m}^2}{0.5} \\
& =5 \times 10^{-4} \mathrm{sm}^2 \text { gram equivalent }{ }^{-1} \\
& \text { We know that } \\
& \frac{\text { Cell constant }}{\mathrm{R}} \\
& \frac{\kappa_2}{\kappa_1}=\frac{R_1}{R_2} \\
& \mathrm{k}_2=\mathrm{k}_1 \times \frac{R_1}{R_2}=2.4 \mathrm{Sm}^{-1} \times \frac{50 \Omega}{480 \Omega}=0.25 \mathrm{Sm}^{-1}
\end{aligned}$

Question 3 .
The emf of the following cell at $25^{\circ} \mathrm{C}$ is equal to $0.34 \mathrm{v}$. Calculate the reduction potential of copper electrode.
$
\operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_2(\mathrm{~g}, 1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \mid \mathrm{Cu}(\mathrm{s})
$
Answer:
SHE Value is zero
$
\begin{aligned}
& \mathrm{E}^{\circ} \text { cell }=\mathrm{E}^{\circ} \mathrm{R}-\mathrm{E}^{\circ} \mathrm{L} \\
& =0.34-0=0.34 \mathrm{~V}
\end{aligned}
$
The reduction potential of copper electrode $=0.34 \mathrm{~V}$
Question 4.
Using the calculated emf value of zinc and copper electrode, calculate the emf of the following cell at $25^{\circ} \mathrm{C}$.
$\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})$
Answer:
$
\begin{aligned}
& \mathrm{E}^{\circ} \text { cell }=\mathrm{E}^{\circ} \mathrm{R}^{-} \mathrm{E}^{\circ} \mathrm{L} \\
& \mathrm{E}_{\mathrm{zn} / \mathrm{zn}^{2+}}=0.76 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{2+}}=0.76 \mathrm{~V} \\
& \mathrm{E}^{\circ} \text { cell }=0.76-(-0.34 \mathrm{~V}) \\
& \mathrm{E}^{\circ} \text { cell }=0.76-(-0.34) \\
& \mathrm{E}^{\circ} \text { cell }=+1.1 \mathrm{~V}
\end{aligned}
$
Question 5.
Write the overall redox reaction which takes place in the galvanic cell, $\mathrm{Pt}(\mathrm{s}) \mid \mathrm{Fe}^{2+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\left\|\mathrm{MnO}_4^{-}(\mathrm{aq}), \mathrm{H}^{+}(\mathrm{aq}), \mathrm{Mn}^{2+}(\mathrm{aq})\right\| \mathrm{Pt}(\mathrm{s})$
Answer:
At Anode half cell $-5 \mathrm{Fe}^{2+}$ (aq) $\rightarrow 5 \mathrm{Fe}^{3+}$ (aq) $+5 \mathrm{e}^{-}$
At cathode half cell $-\mathrm{MnO}_{4(\mathrm{aq})}^{-}+8 \mathrm{H}^{+}{ }_{(\mathrm{aq})}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}_2 \mathrm{O}_{(1)}$
Overall redox reaction $-5 \mathrm{Fe}^{2+}{ }_{(\mathrm{aq})}+\mathrm{MnO}_4^{-}(\mathrm{aq})+8 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 5 \mathrm{Fe}^{3+}{ }_{(\mathrm{aq})}+\mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}_2 \mathrm{O}_{(1)}$
Question 6.
The electrochemical cell reaction of the Daniel cell is
$
\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}_{(\mathrm{s})}
$
What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10 ?
Answer:

$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}_{(\mathrm{s})}$
ln the case $\mathrm{E}^{\circ}$ cell $=1.1 \mathrm{~V}$
Reaction quotient $\mathrm{Q}$ for the above reaction is, $\mathrm{Q}=\frac{\left[\mathrm{zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$
$
\mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}{ }_{\text {cell }}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}
$
If suppose concentration of $\mathrm{Cu}^{2+}$ is $1 . \mathrm{OM}$ then the concentration of $\mathrm{Zn}^{2+}$ is $10 \mathrm{M}$ (why because, ion concentration in the anode compartment increased by a 10 factor)
$
\begin{aligned}
& \mathrm{E}_{\text {cell }}=1.1-\frac{0.0591}{n} \log \left(\frac{10}{1}\right) \\
& =1.1-0.02955 \\
& =1.070 \mathrm{~V} \text { (cell voltage decreased) }
\end{aligned}
$
Thus, the initial voltage is greater than $\mathrm{E}^{\circ}$ because $\mathrm{Q}<1$. As the reaction proceeds, $\left[\mathrm{Zn}^{2+}\right]$ in the anode compartment increases as the zinc electrode dissolves, while $\left[\mathrm{Cu}^{2+}\right]$ in the cathode compartment decreases as metallic copper is deposited on the electrode.
During this process, the $\mathrm{Q}=\left[\mathrm{Zn}^{2+}\right]\left[\mathrm{Cu}^{2+}\right]$ steadily increases and the cell voltage therefore steadily decreases. $\left[\mathrm{Zn}^{2+}\right]$ will continue to increase in the anode compartment and $\left[\mathrm{Cu}^{2+}\right]$ will continue to decrease in the cathode compartment. Thus the value of $\mathrm{Q}$ will increase further leading to a further decrease in value.
Question 7.
A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is $0.783 \mathrm{~g}$. calculate the equivalent mass of the metal.
Answer:
Given,
$I=0.15$ amperes
$t=150$ mins
$=\mathrm{t}=15 \mathrm{O} \times 6 \mathrm{Osec}$
$=\mathrm{t}=9000 \mathrm{sec}$
$\mathrm{Q}=\mathrm{It}$
$=\mathrm{Q}=0.15 \times 9000$ coulombs
$=\mathrm{Q}=1350$ coulombs
Hence, 135 coulombs of electricity deposit is equal to $0.783 \mathrm{~g}$ of metal.
96500 coulombs of electricity, $\frac{0.783 x 96500}{1350}=55.97 \mathrm{gm}$ of metal
Hence equivalent mass of the metal is 55.97