NCERT Chemistry Solutions for Class 12 Chapter 3 – Electrochemistry

Q.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

Ans.

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

Mg, Al, Zn, Fe, Cu

Q.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.

Ans.

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

Q.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.

Ans.

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

${\mathrm{Zn}}_{\left(\mathrm{s}\right)}\parallel {\mathrm{Zn}}_{\left(\mathrm{aq}\right)}^{2+}\parallel {\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}\parallel {\mathrm{Ag}}_{\left(\mathrm{s}\right)}$

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

(ii) Metal ions are carriers of current in the internal circuit and electrons act as current carriers in the external circuit,hence current will flow from silver to zinc.

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

${\mathrm{Zn}}_{\left(\mathrm{s}\right)}\to {\mathrm{Zn}}_{\left(\mathrm{aq}\right)}^{2+}+2_{\mathrm{e}}^{\_}$

The reaction taking place at the cathode is given by,

${\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}{+}_{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{\left(\mathrm{s}\right)}$

Q.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 ∆rG^θ and equilibrium constant of the reactions.

Ans.

Q.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).

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}For the given reaction},\text{\hspace{0.17em}the Nernst equation can be given as}:\\ {\mathrm{E}}_{\mathrm{cell}}=\mathrm{}{\mathrm{E}}_{\mathrm{cell}}^{\varnothing }-\frac{0.0591}{\mathrm{n}}\log \mathrm{}\frac{\left[{\mathrm{Mg}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\\ =\{0.34-(-2.36)\}-\frac{0.0591}{2}\mathrm{}\log \mathrm{}\frac{.001}{.0001}\\ =2.7-\mathrm{}\frac{0.0591}{2}\log \mathrm{}10\\ =2.7-0.02955\\ =2.67\mathrm{}\mathrm{V}\mathrm{}\left(\text{approximately}\right)\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}For the given reaction},\text{\hspace{0.17em}the Nernst equation can be given as}:\\ {\mathrm{E}}_{\mathrm{cell}}=\mathrm{}{\mathrm{E}}_{\mathrm{cell}}^{\varnothing }-\frac{0.0591}{\mathrm{n}}\log \mathrm{}\frac{\left[{\mathrm{Fe}}^{2+}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}\\ =\{0-(-0.44)\}-\frac{0.0591}{2}\mathrm{}\log \mathrm{}\frac{0.001}{1^2}\\ \mathrm{}=0.44-\mathrm{}0.02955\mathrm{}(-3)\\ \mathrm{}=0.52865\mathrm{}\mathrm{V}\\ =0.53\mathrm{}\mathrm{V}\mathrm{}\left(\text{approximately}\right)\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}For the given reaction},\text{\hspace{0.17em}the Nernst equation can be given as}:\\ {\mathrm{E}}_{\mathrm{cell}}=\mathrm{}{\mathrm{E}}_{\mathrm{cell}}^{\varnothing }-\frac{0.0591}{\mathrm{n}}\log \mathrm{}\frac{\left[{\mathrm{Sn}}^{2+}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}\\ =\{0-(-0.14)\}-\frac{0.0591}{2}\mathrm{}\log \mathrm{}\frac{0.050}{{(0.020)}^2}\\ =0.14-\mathrm{}0.0295\mathrm{}\mathrm{x}\mathrm{}\log \mathrm{}125\\ =0.14-0.062\\ =0.078\mathrm{}\mathrm{V}\\ \mathrm{}=0.08\mathrm{}\mathrm{V}\mathrm{}\left(\text{approximately}\right)\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}For the given reaction},\text{\hspace{0.17em}the Nernst equation can be given as}:\\ {\mathrm{E}}_{\mathrm{cell}}=\mathrm{}{\mathrm{E}}_{\mathrm{cell}}^{\varnothing }-\frac{0.0591}{\mathrm{n}}\log \mathrm{}\frac{1}{{\left[{\mathrm{Br}}^{-}\right]}^2{\left[{\mathrm{H}}^{+}\right]}^2}\\ =(0-1.09)-\frac{0.0591}{2}\mathrm{}\log \mathrm{}\frac{1}{{(0.010)}^2{(0.030)}^2}\\ =\mathrm{}-1.09-\mathrm{}0.02955\mathrm{}\times \mathrm{}\log \mathrm{}\frac{1}{0.00000009}\\ =\mathrm{}-1.09-0.02955\mathrm{}\times \mathrm{}\log \mathrm{}\frac{1}{9\mathrm{}\times \mathrm{}{10}^{-8}}\\ =\mathrm{}-1.09-\mathrm{}0.02955\mathrm{}\times \mathrm{}\log \mathrm{}(1.11\mathrm{}\times \mathrm{}{10}^7)\\ =\mathrm{}-1.09-0.02955\mathrm{}(0.0453+7)\\ =\mathrm{}-1.09-0.208\mathrm{}\\ =\mathrm{}-1.298\end{array}$

Q.6 In the button cells widely used in watches and other devices the following reaction takes place.Zn(s)+ Ag2O(s)+ H2O(l)→Zn2+(aq)+ 2Ag(s)+ 2OH−(aq)Determine=ΔrGϕ and Eϕ for the reaction.

Ans.

$\begin{array}{l}{\mathrm{Zn}}_{\left(\mathrm{s}\right)}\to {\mathrm{Zn}}_{\left(\mathrm{aq}\right)}^{2+}+2_{\mathrm{e}}^{-};\mathrm{}{\mathrm{E}}^{\varnothing }=0.76\mathrm{}\mathrm{V}\\ {\mathrm{Ag}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+2_{\mathrm{e}}^{-}\to 2{\mathrm{Ag}}_{\left(\mathrm{s}\right)}+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{{}^{-}};\mathrm{}{\mathrm{E}}^{\varnothing }=0.344\mathrm{}\mathrm{V}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ {\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{Ag}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\to {\mathrm{Zn}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{Ag}}_{\left(\mathrm{s}\right)}+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-};\mathrm{}{\mathrm{E}}^{\varnothing }=1.104\mathrm{}\mathrm{V}\\ \\ \therefore {\mathrm{E}}^{\varnothing }=1.104\mathrm{}\mathrm{V}\\ \text{We know that},\\ {\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{G}}^{\varnothing }=-{\mathrm{nFE}}^{\varnothing }\\ =-2\times \mathrm{}96487\times \mathrm{}1.04\\ =-213043.296\mathrm{}\mathrm{J}\\ =-213.04\mathrm{}\mathrm{kJ}\end{array}$

Q.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Ans.

Conductivity of a solution is defined as the conductance of an electrolyte in which electrodes are unit distance apart and have area of cross section between them as unity. If ρ is resistivity, then we can write:

$\mathrm{\kappa }=\frac{1}{\mathrm{\rho }}$

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.

$\mathrm{i}.\mathrm{e}.,\mathrm{G}=\mathrm{k}\frac{\mathrm{a}}{\mathrm{l}}=\mathrm{k}.1=\mathrm{k}$

(Since a = 1, l = 1)

Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity:

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

${\wedge }_{\mathrm{m}}=\mathrm{k}\frac{\mathrm{A}}{\mathrm{l}}$

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

$\therefore {\wedge }_{\mathrm{m}}=\mathrm{kV}$

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

The variation of ∧_m with √C for strong and weak electrolytes is shown in the following plot:

Q.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.

Ans.

$\begin{array}{l}\begin{array}{l}\mathrm{Given},\\ \mathrm{\kappa }=0.0248\mathrm{S}{\mathrm{cm}}^{-1}\\ \mathrm{c}=0.20\mathrm{M}\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Molar\hspace{0.17em}conductivity},{\wedge }_{\mathrm{m}}=\frac{\text{k\hspace{0.17em}}\times 1000}{\mathrm{c}}\\ =\frac{0.0248\times 1000}{0.2}\\ =124\mathrm{S}c{\mathrm{m}}^2{\mathrm{mol}}^{-1}\end{array}$

Q.9 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M 0.001 0.010 0.020 0.050 0.100

10² × κ/S m^-1 1.237 11.85 23.15 55.53 106.74

Calculate ∧_m for all concentration and draw a plot between ∧_m and c½. Find the value of ∧_m⁰.

Ans.

Given,
κ = 1.237 × 10-2 S m^-1, c = 0.001 M
Then, κ = 1.237 × 10^-4 S cm^-1, c½ = 0.0316 M^1/2
∴∧_m = κ/C
=(1.237 x 10^-4 Scm^-1 x 1000cm³/ 0.001 mol L^-1 x L
=123.7 Scm² L^-1
Given,
κ = 11.85 × 10-2 S m-1, c = 0.010M
Then, κ = 11.85 × 10-4 S cm-1, c½ = 0.1 M^1/2
∴∧_m = κ/C
=118.5 Scm² L^-1
Given,
κ = 23.15 × 10-2 S m-1, c = 0.020 M
Then, κ = 23.15 × 10^-4 S cm^-1, c^1/2 = 0.1414 M^1/2
∴∧_m = κ/CL
=115.8 Scm² L^-1
Given,
K = 55.53 × 10-2 S m^-1, c = 0.050 M
Then, κ = 55.53 × 10^-4 S cm^-1, c^1/2 = 0.2236 M^1/2
∴∧_m = κ/C
=111.1 1 Scm² L^-1
Given,
κ = 106.74 × 10^-2 S m^-1, c = 0.100 M
Then, κ = 106.74 × 10^-4 S cm^-1, c^1/2 = 0.3162 M^1/2
∴∧_m = κ/C
=106.74 Scm² L^-1
Now, we have the following data:

C^1/2 ⁄M^1/2 0.0316 0.1 0.1414 0.2236 0.3162

∧_m (Scm² mol^-1) 123.7 118.5 115.8 111.1 106.74

Q.10 Conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity and if ∧_m⁰ for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?

Ans.

$\begin{array}{l}{\text{Given, κ = 7.896 × 10}}^{\text{–5}}{\text{Scm}}^{\text{–1}}\\ {\text{c = 0.00241 mol L}}^{\text{–1}}\\ \text{Then, molar conductivity,}{\mathrm{\Lambda }}_{\text{m}}\text{=}\frac{\text{K}}{\text{C}}\\ \text{=}\frac{{\text{7.896 × 10}}^{\text{–5}}{\text{Scm}}^{\text{–1}}}{{\text{0.00241 mol L}}^{\text{–1}}}\text{×}\frac{{\text{1000 cm}}^{\text{3}}}{\text{L}}\\ =32.76{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}\\ \mathrm{Again},{\mathrm{\Lambda }}_{\mathrm{m}}^0=390.5{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}\\ \mathrm{\alpha }\text{\hspace{0.17em}=}\frac{{\mathrm{\Lambda }}_{\text{m}}}{{\mathrm{\Lambda }}_{\text{m}}^{\text{0}}}\text{=}\frac{{\text{32.76 Scm}}^{\text{2}}{\text{mol}}^{\text{–1}}}{{\text{390.5 Scm}}^{\text{2}}{\text{mol}}^{\text{–1}}}\\ \mathrm{Now},\\ =0.084\\ \mathrm{Dissociation}\mathrm{}\mathrm{constant},{\mathrm{K}}_a=\frac{{\mathrm{c\alpha }}^2}{(1-\mathrm{\alpha })}\\ \text{=}\frac{\left({\text{0.00241 molL}}^{\text{–1}}\right){\left(\text{0.084}\right)}^{\text{2}}}{\left(\text{1\hspace{0.17em}–\hspace{0.17em}0.084}\right)}\\ =1.86\times {10}^{-5}{\mathrm{molL}}^{-1}\end{array}$

Q.11 

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Ans.

\begin{array}{l}{\text{(i) Al}}^{3+}{\text{+3}}_{\text{e}}^{\text{–}}\to \text{Al}\\ \therefore \text{Required charge = 3F}\\ \text{= 3 × 96487 C}\\ \text{= 289461C}\\ \text{(ii)}{\text{Cu}}^{2+}{\text{+2}}_{\text{e}}^{\text{–}}\to \text{Cu}\\ \therefore \text{Required charge = 2 F = 2 × 96487 C}\\ \text{= 192974C}\\ {\text{(iii) MnO}}_4^{–}\to {\text{Mn}}^{2+}\\ \text{i}\text{.e}{\text{., Mn}}^{7+}{\text{+ 5}}_e^{–}\to {\text{Mn}}^{2+}\\ \text{Required charge = 5 F}\\ \text{= 5 × 96487 C}\\ \text{= 482435C}\end{array}

Q.12 How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl₂.
(ii) 40.0 g of Al from molten Al₂O₃.

Ans.

$\begin{array}{l}\text{(i) According to the question,}\\ {\text{Ca}}^{2+}\text{+ 2e –}\to \text{Ca}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}40g}\\ \text{Electricity required to produce 40g of calcium = 2F}\\ \text{Electricity required to produce 20g of calcium =}\frac{\text{2\hspace{0.17em}×\hspace{0.17em}20}}{\text{40}}\text{F}\\ \text{Therefore, electricity required to produce 20g of calcium = 1F}\\ \text{(ii) According to the question,}\\ {\text{Al}}^{3+}{\text{+ 3}}_{\mathrm{e}}^{–}\to \text{Al}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 27g}\\ \text{Electricity required to produce 27g of Al = 3F}\\ \text{=}\frac{3\times 40}{27}\text{\hspace{0.17em}F}\\ \text{Therefore, electricity required to produce 40g of Al = 4.44 F.}\end{array}$

Q.13

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes.
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Ans.

\begin{array}{l}\left(\text{i}\right)\text{At}\text{cathode:The}\text{following}\text{reduction}\text{reactions}\text{compete}\text{to}\text{take}\text{place}\text{at}\text{the}\text{cathode}\text{.}\\ {\text{Ag}}_{(\text{aq})}^{+}+{\text{e}}^{-}\to {\text{Ag}}_{(\text{s})};{\text{E}}^{°}=0.80\text{V}\\ {\text{H}}_{(\text{aq})}^{+}+{\text{e}}^{-}\to \frac{1}{2}{\text{H}}_{2\left(\text{g}\right)};{\text{E}}^{°}=0.00\text{V}\\ \text{The}\text{reaction}\text{with}\text{a}\text{higher}\text{value}\text{of}{\text{E}}^{\text{°}}\text{takes}\text{place}\text{at}\text{the}\text{cathode}\text{.}\end{array} \begin{array}{l}\text{Therefore,deposition of}\text{silver}\text{will}\text{take}\text{place}\text{at}\text{the}\text{cathode}\text{.}\\ \text{At anode:}\\ \text{The}\text{Ag}\text{anode}\text{is}\text{attacked}\text{by}{\text{NO}}_{\text{3}}^{\text{–}}\text{ions}\text{.Therefore,the}\text{silver}\text{electrode}\\ \text{at}\text{the}\text{anode dissolves in the solution to from Ag}\text{.}\\ \left(\text{ii}\right)\text{At}\text{cathode:The}\text{following}\text{reduction}\text{reactions}\text{compete}\text{to}\text{take}\text{place}\text{at}\text{the}\text{cathode}\\ {\text{Ag}}_{(\text{aq})}^{+}+{\text{e}}^{-}\to {\text{Ag}}_{(\text{s})};{\text{E}}^{°}=0.80\text{V}\\ {\text{H}}_{(\text{aq})}^{+}+{\text{e}}^{-}\to \frac{1}{2}{\text{H}}_{2(\text{g})};{\text{E}}^{°}=0.00\text{V}\\ \text{The}\text{reaction}\text{with}\text{a}\text{higher}\text{value}\text{of}{\text{E}}^{\text{°}}\text{takes}\text{place}\text{at}\text{the}\text{cathode}\text{.}\end{array} $\begin{array}{l}\text{At}\text{cathode}:\\ \text{The\hspace{0.17em}following\hspace{0.17em}oxidation\hspace{0.17em}reactions\hspace{0.17em}are\hspace{0.17em}possible\hspace{0.17em}at\hspace{0.17em}the\hspace{0.17em}anode}.\\ {\text{Cl}}_{\left(\text{aq}\right)}^{-}\to \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\text{Cl}}_{2\left(\text{g}\right)}+{\text{e}}^{-1};{\text{E}}^{\mathrm{°}}=1.36\text{V}\\ 2{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\to {\text{O}}_{2\left(\text{g}\right)}+4{\text{H}}_{\left(\text{aq}\right)}^{+}+4{\text{e}}^{-};{\text{E}}^{\mathrm{°}}=1.23\text{V}\\ \mathrm{At}\mathrm{the}\mathrm{anode},\mathrm{there}\mathrm{action}\mathrm{with}\mathrm{a}\mathrm{lower}\mathrm{value}\mathrm{of}{\mathrm{E}}^{\text{°}}\mathrm{is}\mathrm{preferred}.\end{array}$ \begin{array}{l}\text{But}\text{due}\text{to}\text{the}\text{overpotential of oxygen,}\\ {\text{Cl}}^{\text{–}}{\text{gets oxidised at the anode to produce Cl}}_{\text{2}}\text{gas}\text{.}\end{array}

Q.14 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe³⁺(aq) and I^– (aq)
(ii) Ag⁺ (aq) and Cu(s)
(iii) Fe³⁺ (aq) and Br^‑ (aq)
(iv) Ag(s) and Fe³⁺ (aq)
(v) Br₂ (aq) and Fe²⁺ (aq)

Ans.

$\begin{array}{l}\left(\mathrm{i}\right){\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\times 2;\mathrm{E}\mathrm{°}=+0.77\mathrm{V}\\ 2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{I}}_2\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\mathrm{E}\mathrm{°}=-0.54\mathrm{V}\\ \stackrel{--------------------------------------------------------------}{2{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{s}\right);\mathrm{E}\mathrm{°}=+0.23\mathrm{V}}\\ {\text{Since E}}^{\mathrm{°}}\text{for the overall reaction is positive},\\ {\text{the reaction between Fe}}^{\text{3}+}{\left(\text{aq}\right)}_{}{\text{and I}}^{–}\left(\text{aq}\right)\text{is feasible}.\\ \left(\mathrm{ii}\right){\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{Ag}\left(\mathrm{s}\right)\times 2;\mathrm{E}\mathrm{°}=+0.80\mathrm{V}\\ \mathrm{Cu}\left(\mathrm{s}\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\mathrm{E}\mathrm{°}=-0.34\mathrm{V}\\ \stackrel{--------------------------------------------------------------------}{\begin{array}{l}2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)\to 2\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right);\mathrm{E}\mathrm{°}=+0.46\mathrm{V}\\ \end{array}}\\ \text{Since E}\mathrm{°}\text{for the overall reaction is positive},\\ {\text{the reaction between Ag}}^{+}{\left(\text{aq}\right)}^{}\text{and Cu}\left(\text{s}\right)\text{is feasible}.\\ \left(\mathrm{iii}\right){\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\times 2;\mathrm{E}\mathrm{°}=+0.77\mathrm{V}\\ 2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Br}}_2\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}\mathrm{E}\mathrm{°}=-1.09\mathrm{V}\\ \stackrel{-------------------------------------------------------------------}{\begin{array}{l}2{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{Br}}_2\left(\mathrm{l}\right);\mathrm{E}\mathrm{°}=-0.32\mathrm{V}\\ \end{array}}\\ \text{Since E}\mathrm{°}\text{for the overall reaction is negative},\\ {\text{the reaction between Fe}}^{\text{3}+}\left(\text{aq}\right){\text{and Br}}^{–}\left(\text{aq}\right)\text{is not feasible}.\\ \left(\mathrm{iv}\right)\mathrm{Ag}\left(\mathrm{s}\right)\to {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\mathrm{E}\mathrm{°}=-0.80\mathrm{V}\\ {\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\mathrm{E}\mathrm{°}=+0.77\mathrm{V}\\ \stackrel{-------------------------------------------------------------------}{\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)\to {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\mathrm{E}\mathrm{°}=-0.03\mathrm{V}}\\ \mathrm{Since}{\mathrm{E}}^{\text{°}}\mathrm{for}\mathrm{the}\mathrm{overall}\mathrm{reaction}\mathrm{is}\mathrm{negative},\mathrm{the}\mathrm{reaction}\mathrm{between}\mathrm{Ag}\left(\mathrm{s}\right)\mathrm{and}\mathrm{Fe}3+\left(\mathrm{aq}\right)\mathrm{is}\mathrm{not}\mathrm{feasible}.\end{array}$ \begin{array}{l}(iv)Ag(s)\to A{g}^{+}(aq)+e^{-}E°=-0.80V\\ F{e}^{3+}(aq)+e^{-}\to F{e}^{2+}(aq)E°=+0.77V\\ \stackrel{-------------------------------------------------------------------}{Ag(s)+F{e}^{3+}(aq)\to A{g}^{+}(aq)+F{e}^{2+}(aq)E°=-0.03V}\\ \text{Since E}°\text{for}\text{the}\text{over}\text{all}\text{reaction}\text{is}\text{positive},\text{the}\text{reaction}\text{between}{\text{Br}}_{\text{2}}\left(\text{aq}\right)\text{and}{\text{Fe}}^{\text{2}+}\left(\text{aq}\right)\text{is feasible}.\end{array}

Q.15 Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Ans.

$\begin{array}{l}\text{According\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}reaction}:\\ {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{Ag}\left(\mathrm{s}\right)\\ 108\mathrm{g}\\ \text{i}.\text{e}.,\text{1}0\text{8g\hspace{0.17em}of\hspace{0.17em}Ag\hspace{0.17em}is\hspace{0.17em}deposited\hspace{0.17em}by\hspace{0.17em}96487\hspace{0.17em}C}.\\ =\frac{96487\times 1.45}{108}=1295.43\mathrm{C}\\ \text{Given},\\ \text{Current}=\mathrm{}\text{1}.\text{5A}\\ =\frac{1295.43}{1.5}\text{s}\\ \mathrm{}\text{Time}\\ =\mathrm{}\text{863}.\text{6s}\\ =\mathrm{}\text{864s}\\ =\mathrm{}\text{14}.\text{4}0\text{\hspace{0.17em}min}\\ \text{Again},\\ {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)\\ 63.5\mathrm{g}\\ \text{i}.\text{e}.,\text{2}\times \text{96487\hspace{0.17em}C\hspace{0.17em}of\hspace{0.17em}charge\hspace{0.17em}deposit}=\text{63}.\text{5g\hspace{0.17em}of\hspace{0.17em}Cu}\\ \text{Therefore},\text{1295}.\text{43C\hspace{0.17em}of\hspace{0.17em}charge\hspace{0.17em}will\hspace{0.17em}deposit}\\ =\frac{63.5\text{x}1295.43}{2\text{x}96487}\text{g}\\ =\mathrm{}0.\text{426g\hspace{0.17em}of\hspace{0.17em}Cu}\\ {\text{Zn}}_{\left(\text{aq}\right)}^{2+}+2_{\text{e}}^{-}\to {\text{Zn}}_{\left(\text{s}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}65}.\text{4 g}\\ \text{i}.\text{e}.,\mathrm{}\text{\hspace{0.17em}2}\times \text{96487C\hspace{0.17em}of\hspace{0.17em}charge\hspace{0.17em}deposit}=\mathrm{}\text{65}.\text{4g\hspace{0.17em}of\hspace{0.17em}Zn}\\ \text{Therefore},\text{1295}.\text{43C\hspace{0.17em}of\hspace{0.17em}charge\hspace{0.17em}will\hspace{0.17em}deposit}\\ =\frac{65.4\text{x}1295.43}{2\text{x}96487}\text{g}\\ =\mathrm{}0.\text{439g\hspace{0.17em}of\hspace{0.17em}Zn}\end{array}$

Q.16 A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Ans.

$\begin{array}{l}\mathrm{Given},\\ \mathrm{Current}=5\mathrm{A}\\ \mathrm{Time}=20\times 60=1200\mathrm{s}\\ \therefore \mathrm{Charge}=\mathrm{Current}\times \mathrm{Time}\\ =5\times 1200\\ =6000\mathrm{C}\\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{reaction},\\ {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Ni}\left(\mathrm{s}\right)\\ 58.7\mathrm{g}\\ \mathrm{Nickel}\mathrm{deposited}\mathrm{by}2\times 96487\mathrm{C}=58.71\mathrm{g}\\ \mathrm{Therefore},\mathrm{nickel}\mathrm{deposited}\mathrm{by}6000\mathrm{C}=\frac{58.71\times 6000}{2\times 96487}\mathrm{g}\\ =1.825\mathrm{g}\\ \mathrm{Hence},1.825\mathrm{g}\mathrm{of}\mathrm{nickel}\mathrm{will}\mathrm{be}\mathrm{deposited}\mathrm{at}\mathrm{the}\mathrm{cathode}.\end{array}$

Q.17 How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H₂O to O₂.
(ii) 1 mol of FeO to Fe₂O₃

Ans.

$\begin{array}{l}\left(\text{i}\right)\mathrm{}\text{According\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}question},\\ {\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\\ \text{Now},\text{we}\mathrm{}\text{can}\text{write}:\\ {\text{O}}^{2-}\to \frac{1}{2}{\text{O}}_2+2_{\text{e}}^{-}\\ \text{Electricity\hspace{0.17em}required\hspace{0.17em}for\hspace{0.17em}the\hspace{0.17em}oxidation\hspace{0.17em}of}\\ {\text{1\hspace{0.17em}mol\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}{\text{O\hspace{0.17em}to\hspace{0.17em}O}}_{\text{2}}\mathrm{}=\mathrm{}\text{2F}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\times \text{96487C}\mathrm{}\\ \mathrm{}=\mathrm{}\text{192974C}\\ \left(\text{ii}\right)\mathrm{}\text{According\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}question},\\ {\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+{\text{e}}^{-1}\\ \text{Electricity\hspace{0.17em}required\hspace{0.17em}for\hspace{0.17em}the\hspace{0.17em}oxidation\hspace{0.17em}of}\\ {\text{1mol of FeO to Fe}}_{\text{2}}{\text{O}}_{\text{3}}=\mathrm{}\text{1F}=\mathrm{}\text{96487C}\end{array}$

Q.18 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^–3 S cm^–1.

Ans.

Given,

Conductivity, κ = 0.146 × 10^–3 S cm^–1

Resistance, R = 1500

Cell constant = κ × R

= 0.146 × 10^–3 × 1500

= 0.219 cm^–1