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Page No 92: - Chapter 3 Electrochemistry Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


Question 3.1:

Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn

Answer:

The following is the order in which the given metals displace each other from the solution of their salts.

Mg, Al, Zn, Fe, Cu

Question 3.2:

Given the standard electrode potentials,

K+/K = −2.93V, Ag+/Ag = 0.80V,

Hg2+/Hg = 0.79V

Mg2+/Mg = −2.37 V, Cr3+/Cr = − 0.74V

Arrange these metals in their increasing order of reducing power.

Answer:

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

Question 3.3:

Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show:

(i) Which of the electrode is negatively charged?

(ii) The carriers of the current in the cell.

(iii) Individual reaction at each electrode.

Answer:

The galvanic cell in which the given reaction takes place is depicted as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6268/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_6c079f9f.gif

(i) Zn electrode (anode) is negatively charged.

(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.

(iii) The reaction taking place at the anode is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6268/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_6705d15b.gif

The reaction taking place at the cathode is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6268/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m7f4882d3.gif

Question 3.4:

Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd

(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

Calculate the ΔrGθ and equilibrium constant of the reactions.

Answer:

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m143279af.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_34e81184.gif

The galvanic cell of the given reaction is depicted as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_10855683.gif

Now, the standard cell potential is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m50224e6f.gif

In the given equation,

n = 6

F = 96487 C mol−1

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_410aa954.gif  = +0.34 V

Then, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7a45b48.gif  = −6 × 96487 C mol−1 × 0.34 V

= −196833.48 CV mol−1

= −196833.48 J mol−1

= −196.83 kJ mol−1

Again,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7a45b48.gif  = −RT ln K

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_744969b7.gif

= 34.496

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif  K = antilog (34.496)

= 3.13 × 1034

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m66611e6b.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m2dbfb9b.gif

The galvanic cell of the given reaction is depicted as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m18dcc2c1.gif

Now, the standard cell potential is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m24e0caff.gif

Here, n = 1.

Then, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m6bf1e40d.gif

= −1 × 96487 C mol−1 × 0.03 V

= −2894.61 J mol−1

= −2.89 kJ mol−1

Again, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m6f667df6.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_161ec6fc.gif

= 0.5073

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6270/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif K = antilog (0.5073)

= 3.2 (approximately)

Question 3.5:

Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s) | Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s)

(ii) Fe(s) | Fe2+(0.001M) || H+(1M)|H2(g)(1bar) | Pt(s)

(iii) Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) | Pt(s)

(iv) Pt(s) | Br2(l) | Br(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s).

Answer:

(i) For the given reaction, the Nernst equation can be given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6271/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_3c8cd696.gif

= 2.7 − 0.02955

= 2.67 V (approximately)

(ii) For the given reaction, the Nernst equation can be given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6271/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m70ae3c6a.gif

= 0.52865 V

= 0.53 V (approximately)

(iii) For the given reaction, the Nernst equation can be given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6271/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_55b13ecb.gif

= 0.14 − 0.0295 × log125

= 0.14 − 0.062

= 0.078 V

= 0.08 V (approximately)

(iv) For the given reaction, the Nernst equation can be given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6271/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m545ab3ff.gif

Also Read : Page-No-93:-Chapter-3-Electrochemistry-Exercise-Solutions-class-12-ncert-solutions-Chemistry

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