Example Problems - Chapter 9 - Electro Chemistry - 12th Chemistry Guide Samacheer Kalvi Solutions

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Example Problems
Question 1.
A conductivity cell has two platinum electrodes separated by a distance $1.5 \mathrm{~cm}$ and the cross sectional area of each electrode is $4.5 \mathrm{sq} \mathrm{cm}$. Using this cell, the resistance of $0.5 \mathrm{~N}$ electrolytic solution was measured as 15 ?. Find the specific conductance of the solution.
Solution.
$
\begin{aligned}
& \mathrm{k}=\frac{1}{15 \Omega} \times \frac{1.5 \times 10^{-2} \mathrm{~m}}{4.5 \times 10^{-4} \mathrm{~m}^2}=2.22 \mathrm{~S} \mathrm{~m}^{-1} \\
& 1=1.5 \mathrm{~cm}=1.5 \times 10^2 \mathrm{~m} \\
& \mathrm{~A}=4.5 \mathrm{~cm}^2 \times\left(10^{-4}\right) \mathrm{m}^2 \\
& \mathrm{R}=15 \Omega
\end{aligned}
$
Question 2.
Calculate the molar conductance of $0.025 \mathrm{M}$ aqueous solution of calcium chloride at $25^{\circ} \mathrm{C}$. The specific conductance of calcium chloride is $12.04 \times 10^2 \mathrm{~S} \mathrm{~m}^3$.
Answer:
Molar conductance
$
\begin{aligned}
& \Lambda \mathrm{m}=\frac{\left(\mathrm{Sm}^{-1}\right) \times 10^{-3}}{\mathrm{M}} \mathrm{mol}^{-1} \mathrm{~m}^3 \\
& =\frac{\left(12.04 \times 10^{-2} \mathrm{Sm}^{-1}\right) \times 10^{-3}}{0.025} \mathrm{~mol}^{-1} \mathrm{~m}^3=581.610^{-2} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^1
\end{aligned}
$

Question 3.
The resistance of a conductivity cell is measured as $190 \Omega$ using $0.1 \mathrm{M} \mathrm{KCI}$ solution (specific conductance of $0.1 \mathrm{M} \mathrm{KCI}$ is $1.3 \mathrm{Sm}^{-1}$ ). When the same cell is filled with $0.003 \mathrm{M}$ sodium chloride solution, the measured resistance is $6.3 \mathrm{~K}$ ? Both these measurements are made at a particular temperature. Calculate the specific and molar conductance of $\mathrm{NaCl}$ solution.
Answer:
Given that
$
\begin{aligned}
& \mathrm{K}=1.3 \mathrm{~S} \mathrm{~m}^{-1} \text { (for } 0.1 \mathrm{M} \mathrm{KCI} \text { solution) } \\
& \mathrm{R}=190 \Omega \\
& \left(\frac{l}{A}\right)=\mathrm{k} \cdot \mathrm{R}=\left(1.3 \mathrm{~S} \mathrm{~m}^{-1}\right)(190 ?)=247 \mathrm{~m}^{-1} \\
& \mathrm{k}_{(\mathrm{NaCl})}=\frac{1}{\mathrm{R}_{(\mathrm{NaCT})}}\left(\frac{l}{A}\right)=\frac{1}{6.3 \mathrm{~K} \Omega}\left(247 \mathrm{~m}^{-1}\right) \\
& =39.2 \times 10^{-3} \mathrm{Sm}^{-1} \\
& \Lambda \mathrm{m}=\frac{\kappa \times 10^{-3} \mathrm{~mol}^{-1} \mathrm{~m}^3}{\mathrm{M}}=\frac{39.2 \times 10^{-3}\left(\mathrm{Sm}^{-1}\right) \times 10^{-3}\left(\mathrm{~mol}^{-1} \mathrm{~m}^3\right)}{0.003} \\
& \Lambda \mathrm{m}=13.04 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$

Question 4.
The net redox reaction of a galvanic cell is given below
$
2 \mathrm{Cr}_{(\mathrm{s})}+3 \mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}{ }_{(\mathrm{aq})}+3 \mathrm{Cu}_{(\mathrm{s})}
$
Write the half reactions and describe the cell using cell notation.
Answer:
Anodic oxidation: $2 \mathrm{Cr}_{(\mathrm{s})} \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+6 \mathrm{e}^{-}$
Cathodic reduction: $3 \mathrm{Cu}^{2+}+6 \mathrm{e}^{-} \rightarrow 3 \mathrm{Cu}_{(\mathrm{s})}$
Cell Notation is: $\mathrm{Cr}_{(\mathrm{s})} \mid \mathrm{CI}^{3+}$ (aq) $\| \mathrm{Cu}^{2+}{ }_{(\mathrm{aq})} \mid \mathrm{Cu}_{(\mathrm{s})}$
Question 5.
Let us calculate the emf of the following cell at $25^{\circ} \mathrm{C}$ using Nernst equation.
$\mathrm{Cu}(\mathrm{s}) \mid \mathrm{Cu}^{2+}(0.25 \mathrm{aq}, \mathrm{M}) \| \mathrm{Fe}^{3+}\left(0.05\right.$ aq M) || $\mathrm{Fe}^{2+}(0.1 \mathrm{aq} \mathrm{M}) \mid \mathrm{pt}(\mathrm{s})$
Answer:
Given: $\left(\mathrm{E}^{\circ}\right)_{\mathrm{Fe}^{3+}} \mid \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$ and $\left(\mathrm{E}^{\circ}\right)_{\mathrm{Cu} 2+\mathrm{Cu}}=0.34 \mathrm{~V}$
Half reactions are
$
\begin{aligned}
& \mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \ldots \ldots \\
& 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})
\end{aligned}
$
the overall reaction is $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Fe}^{2+}(\mathrm{aq})$, and $\mathrm{n}=2$ Apply Nernst equation at $25^{\circ} \mathrm{C}$

$
\begin{aligned}
& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{Fe}^{2+}\right]^2}{\left[\mathrm{Fe}^{3+}\right]^2} \quad[\because \mathrm{Cu}(\mathrm{S})=1] \\
& \mathrm{E}_{\mathrm{cell}}^{\circ}=\left(\mathrm{E}_{\mathrm{ox}}^{\circ}\right)_{\mathrm{Cu} \mid \mathrm{Cu}^{2+}}+\left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}
\end{aligned}
$
Given standard reduction potetial of $\mathrm{Cu}^{2+} \mid \mathrm{Cu}$ is 0.34
$
\begin{aligned}
& \therefore\left(\mathrm{E}_{\mathrm{ox}}^{\circ}\right)_{\mathrm{Cu} \mid \mathrm{Cu}^{2+}}=-0.34 \mathrm{~V} \\
& \left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)_{\mathrm{Fe}^{3+}} \mathrm{Fe}^{2+}=0.77 \mathrm{~V} \\
& \therefore \mathrm{E}_{\text {cell }}^{\circ}=-0.34+0.77=0.43 \mathrm{~V}
\end{aligned}
$
$
\begin{aligned}
\therefore \mathrm{E}_{\text {cell }} & =0.43-\frac{0.0591}{2} \times \log \frac{(0.25)(0.1)^2}{(0.005)^2} \\
& =0.43-\frac{0.0591}{2} \times 2 \\
& =0.43-0.0591 \\
& =0.3709 \mathrm{~V} \\
& =\log \frac{(0.25)(0.1)^2}{(0.005)^2} \\
& =\log \frac{25 \times 10^{-2} \times 1 \times 10^{-2}}{25 \times 10^{-6}} \\
& =\log 10^2 \\
& =2 \log 10=2 .
\end{aligned}
$

Question 6.
A solution of silver nitrate Is electrolysed for 20 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode.
Answer:
Electrochemical reaction at cathode is $\mathrm{Ag}^{+}+\mathrm{e}^{-} \mathrm{Ag}$ (reduction)
$\mathrm{m}=\mathrm{ZIt}$
$\mathrm{m}=\frac{108 \mathrm{gmol}^{-1}}{96500 \mathrm{Cmol}^{-1}} \times 2400 \mathrm{C}$
$\mathrm{m}=2.68 \mathrm{~g}$
$\mathrm{z}=\frac{\text { molarmass of Ag }}{(96500)}=\frac{108}{1 \times 96500}$
$\mathrm{I}=2 \mathrm{Ag}$
$\mathrm{t}=20 \times 60 \mathrm{~s}=1200 \mathrm{~s}$
It $=2 \mathrm{~A} \times 1200 \mathrm{~S}=2400 \mathrm{C}$