Electrochemistry

When copper sulphate ($ \mathrm{CuSO}_4 $) is dissolved in water, it splits into ions as follows:

$ \mathrm{CuSO}_4 \longrightarrow \mathrm{Cu}^{2+} + \mathrm{SO}_4^{2-} $

During electrolysis, copper ions move towards the cathode (negative electrode) where they gain electrons and get deposited as solid copper metal:

$ \mathrm{Cu}^{2+} + 2e^- \longrightarrow \mathrm{Cu(s)} $

According to Faraday’s laws of electrolysis, the mass of a substance deposited is directly proportional to the quantity of charge passed. The mathematical form is:

$ W = \frac{E \times I \times t}{96500} $

where,

$ W $ = mass of substance deposited, $ E $ = electrochemical equivalent (equal to molar mass / number of electrons involved), $ I $ = current in amperes, $ t $ = time in seconds, and $ 96500\, \mathrm{C/mol} $ = 1 Faraday.

For copper, one atom needs 2 electrons, so $ E = \frac{63}{2} $.

Now substituting the given values: $ I = 1.5\,\mathrm{A} $, $ t = 10 \times 60 = 600\,\mathrm{s} $

$ W = \frac{63 \times 1.5 \times 600}{2 \times 96500} $

Simplifying,

$ W = \frac{56,700}{193000} = 0.2938\,\mathrm{g} $

Hence, the mass of copper deposited on the cathode is $ 0.2938\,\mathrm{g} \approx 0.29\,\mathrm{g} $.

$ \begin{aligned} & \text { } \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{H}^{\oplus}(\mathrm{aq})+2 \mathrm{e}^{\ominus} \\ & \because E=\mathrm{E}^{\circ}-\frac{2.303 R \mathrm{~T}}{\mathrm{nF}} \log \frac{\left[\mathrm{H}^{+}\right]^2}{\mathrm{P}_{\mathrm{H}_2}} \\ & {\left[\mathrm{H}^{+}\right]=0.02 \mathrm{M}(\text { from } \mathrm{HCl})} \\ & P_{\mathrm{H}_2}=2 \mathrm{~atm} \\ & \frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.059 \\ & \mathrm{E}=0-\frac{0.059}{2} \log \frac{(0.02)^2}{2} \\ & =-0.0295 \log \left(\frac{0.0004}{2}\right) \\ & \mathrm{E}=-0.0295 \log \left(2 \times 10^{-4}\right) \\ & =-0.0295\left(\log 2+\log 10^{-4}\right) \\ & =-0.0295[0.301-4]=0.109 \mathrm{~V} \end{aligned} $

$ \begin{aligned} & \Delta \mathrm{G}^{\circ} \mathrm{Fe}^{3+} / \mathrm{Fe}=\Delta \mathrm{G}^{\circ} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}+\Delta \mathrm{G}^{\circ} \mathrm{Fe}^{2+} / \mathrm{Fe} \\ & -\mathrm{nFE} \mathrm{E}^{\circ}=-\mathrm{nFE} \mathrm{E}^{\circ}+-\mathrm{nFE} \mathrm{~F}^{\circ} \\ & -3 \mathrm{~F}(-0.04)=-1 \mathrm{FE}^{\circ}+2 \mathrm{~F}(0.44) \\ & 3 \mathrm{~F}(0.04)=-\mathrm{Fe}^{\circ}+0.88 \mathrm{~F} \\ & 0.12=-\mathrm{E}^{\circ}+0.88 \\ & 0.12-0.88=-\mathrm{E}^{\circ} \\ & -0.76=-\mathrm{E}^{\circ} \\ & \mathrm{E}^{\circ}=0.76 \mathrm{~V} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & y=m x+c \\ & \text { slope }=-90 \end{aligned}\\ &\text { According to } \lambda_{\mathrm{m}}=\lambda_{\mathrm{m}}^0-\mathrm{A} \sqrt{\mathrm{c}}\\ &\begin{aligned} & 145=\lambda_{\mathrm{m}}^0-90 \times 0.1 \\ & \lambda_{\mathrm{m}}^{\circ}=154 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{m}(\mathrm{XY})}^{\circ}=\lambda_{\mathrm{X}^{+}}^{\circ}+\lambda_{\mathrm{Y}^{-}}^{\circ} \\ & 154=74+\lambda_{\mathrm{Y}^{-}}^{\circ} \\ & \lambda_{\mathrm{Y}^{-}}^{\circ}=80 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $

Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion.

$\bullet$ Production of H$_2$ by electrolysis of water is an example of electrolytic cell.

To find the standard Gibbs energy change for the reaction, we can use the relationship between the standard Gibbs free energy change ($\Delta G^\circ$) and the standard cell potential ($E^\circ$) given by the following equation:

$\Delta G^\circ = -nFE^\circ$

Where:

• $n$ is the number of moles of electrons transferred in the reaction.
• $F$ is the Faraday constant ($1 \mathrm{~F} = 96487 \mathrm{~C \cdot mol^{-1}}$).
• $E^\circ$ is the standard cell potential.

From the given reaction:

$\mathrm{Zn}(\mathrm{s}) + \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{Fe}(\mathrm{s})$

It can be observed that $n = 2$ because two electrons are transferred from zinc to iron.

Given data:

• $E^\circ = 0.32 \mathrm{~V}$
• $1 \mathrm{~F} = 96487 \mathrm{C/mol}$

Now, substitute the values into the equation:

$\Delta G^\circ = -nFE^\circ$

$\Delta G^\circ = -2 \times 96487 \mathrm{~C/mol} \times 0.32 \mathrm{~V}$

Calculate the value:

$\Delta G^\circ = -2 \times 96487 \mathrm{~C/mol} \times 0.32 \mathrm{~V}$

$\Delta G^\circ = -2 \times 30875.84 \mathrm{~J/mol}$

$\Delta G^\circ = -61751.68 \mathrm{~J/mol}$

Converting the units to kJ/mol:

$\Delta G^\circ = -61.75168 \mathrm{~kJ/mol}$

The closest answer, when rounded to two decimal places, is:

$\Delta G^\circ = -61.75 \mathrm{~kJ/mol}$

Therefore, the correct option is:

Option A: $-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$

To answer this question, we need to calculate the number of Faraday's required for each of the reactions in List I.

Reaction A: Conversion of H₂O to O₂

The reaction for electrolysis of water to produce oxygen can be written as:

$\text{2H}_2\text{O (l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^-$

Each mole of O₂ requires the transfer of 4 moles of electrons. Therefore, to produce 1 mole of O₂, you need 4 Faraday's of charge. However, since the reaction shown in the table suggests the production of only 1 mole, it would require 2 Faraday's. This is because we are dealing with a half reaction in which only 2 moles of electrons are required for every mole of O₂ produced from 1 mole of H₂O.

Reaction B: Conversion of MnO₄^- to Mn²⁺

The reduction reaction for permanganate to manganese ion is:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

This reaction indicates that 5 moles of electrons are involved in the reduction of 1 mole of MnO₄^- to Mn²⁺. Therefore, 5 Faraday's are required.

Reaction C: Production of 1.5 moles of Ca from molten CaCl₂

The electrolytic production of calcium from its chloride can be illustrated as follows:

$\text{Ca}^{2+} + 2\text{e}^- \rightarrow \text{Ca}$

Each mole of Ca produced requires 2 Faraday's of electrons considering the transfer of 2 moles of electrons. For 1.5 moles of Ca, the total Faraday's required would be 3 Faraday's, because, 1.5 * 2 = 3F.

Reaction D: Oxidation of FeO to Fe₂O₃

The reaction can be considered in terms of iron oxidation states:

$\text{4FeO} \rightarrow \text{2Fe}_2\text{O}_3 + \text{O}_2 + 4\text{e}^-$

Each mole of O₂ requires 2 Faraday's of charge. If the products include 2 moles of Fe₂O₃ and 1 mole of O₂, thus, only 2 Faraday's will be required for 1 mole of FeO to react (considering FeO as just a part of the full reaction).

Correct Matching:

A-II, B-IV, C-I, D-III

The mappings are A-II (2F for H₂O to O₂), B-IV (5F for MnO₄^- to Mn²⁺), C-I (3F for 1.5 mol of Ca from CaCl₂), and D-III (2F for FeO to Fe₂O₃).

Therefore, the correct answer is Option A:

A-II, B-IV, C-I, D-III.

The mass of copper deposited when passing an electric current through a copper sulphate solution can be calculated using Faraday's laws of electrolysis. The first law states that the amount of a substance deposited or liberated at an electrode during electrolysis is proportional to the amount of electricity (charge) passed through the electrolyte.

To find the mass of copper deposited, we use the formula:

$ m = \frac{M \times I \times t}{n \times F} $

Where:

• $ m $ is the mass of the substance deposited (in grams),


• $ M $ is the molar mass of the substance (in g/mol),


• $ I $ is the current (in amperes, A),


• $ t $ is the time electricity is passed through the solution (in seconds),


• $ n $ is the number of moles of electrons required to deposit or dissolve 1 mole of the substance (valence number, which is 2 for copper in copper sulphate solution as copper ions are $\mathrm{Cu^{2+}}$),


• $ F $ is the Faraday constant, approximately $ 96487 \mathrm{C/mol} $ of electrons.

Given:

• $ M = 63 \mathrm{~g/mol} $


• $ I = 9.6487 \mathrm{~A} $


• $ t = 100 \mathrm{~s} $


• $ n = 2 $


• $ F = 96487 \mathrm{~C/mol} $

Substitute these values into the formula:

$ m = \frac{63 \mathrm{~g/mol} \times 9.6487 \mathrm{~A} \times 100 \mathrm{~s}}{2 \times 96487 \mathrm{~C/mol}} $

Calculate the value of $ m $:

$ m = \frac{63 \times 9.6487 \times 100}{2 \times 96487} $

$ m = \frac{60828.21}{192974} $

$ m \approx 0.315 \mathrm{~g} $

Hence, the mass of copper deposited is approximately 0.315 g, making Option B the correct answer.

$\mathrm{Tl}^3$ act as an oxidising agent not reducing agent.

B and D statement are correct.

$\begin{aligned} \mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\mathrm{C}}^{\mathrm{o}}-\mathrm{E}_{\mathrm{A}}^{\mathrm{o}} \\ & =(1.33)-(-0.44) \\ & =+1.77 \mathrm{~V} \end{aligned}$

Centimolar solution $=\frac{1}{100} \mathrm{M}=0.01 ~\mathrm{M}$

Conductivity $(k)=0.0210 ~\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$

Resistance $(\mathrm{R})=60 ~\mathrm{ohm}$

$k=\frac{1}{\mathrm{R}}\left(\frac{\ell}{\mathrm{A}}\right)$

$\Rightarrow 0.0210=\frac{1}{60}\left(\frac{\ell}{\mathrm{A}}\right) \Rightarrow \frac{\ell}{\mathrm{A}}=1.26 \mathrm{~cm}^{-1}$

$\mathrm{\Delta_rG=-nFE_{cell}}$

$\mathrm{E_{cell}}$ is an intensive property and $\mathrm{\Delta_r G}$ is an extensive property as it depends on number of $\mathrm{{e^\Theta }}$ transferred in cell reaction.

Since $E_{OP}^0$ of Al is more than Co²⁺, so at anode Al will oxidise and at cathode Co³⁺ will reduce.

$E_{Cell}^0 = {(E_{Cathode}^0)_{RP}} - {(E_{Anode}^0)_{RP}}$

$ = E_{C{o^{3 + }}/C{o^{2 + }}}^0 - E_{A{l^{3 + }}/Al}^0$

$ = (1.81) - ( - 1.66)$

$ = + 3.47\,V$

Zn(s) + Cu²⁺(aq) $\to$ Zn²⁺(aq) + Cu(s)

E$_{cell}^0$ = 1.1 V

$\Delta$G$^\circ$ = $-$nFE$_{cell}^0$

$\therefore$ n = 2

$\Delta$G$^\circ$ = $-$2 $\times$ 96487 $\times$ 1.1

$\Delta$G$^\circ$ = $-$212271.4 J mol^$-$1

$\Delta$G$^\circ$ = $-$212.27 kJ mol^$-$1

For a reaction to be spontaneous, E$_{cell}^o$ must be positive.

$\bullet$ For, FeSO₄(aq) + Zn(s) $\to$ ZnSO₄(aq) + Fe(s)

E$_{cell}^o$ = E$_{cathode}^o$ $-$ E$_{anode}^o$

= $-$0.44 V $-$ ($-$0.76 V)

= 0.32 V

$\bullet$ For, 2CuSO₄(aq) + 2Ag(s) $\to$ 2Cu(s) + Ag₂SO₄(aq)

E$_{cell}^o$ = 0.34 V $-$ 0.80 V

= $-$0.46 V

$\bullet$ For, CuSO₄(aq) + Zn(s) $\to$ ZnSO₄(aq) + Cu(s)

E$_{cell}^o$ = 0.34 V $-$ ($-$0.76 V)

= 1.1 V

$\bullet$ For, CuSO₄(aq) + Fe(s) $\to$ FeSO₄(aq) + Cu(s)

E$_{cell}^o$ = 0.80 V $-$ ($-$0.44 V)

= 1.24 V

$\bullet$ $MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$ ..... (i)

$E_{MnO_4^ - /M{n^{2 + }}}^o = - E_{M{n^{2 + }}/MnO_4^ - }^o = 1.51\,V$

$\bullet$ ${H_2}O \to {1 \over 2}{O_2} + 2{H^ + } + 2{e^ - }$ ..... (ii)

$E_{{O_2}/{H_2}O}^o = 1.223\,V$

Using 2 $\times$ (i) + 5 $\times$ (ii), net cell reactions is

$2MnO_4^ - + 6{H^ + } \to 2M{n^{2 + }} + {5 \over 2}{O_2} + 3{H_2}O$

$E_{cell}^o = E_C^o - E_A^o = E_{MnO_4^ - /M{n^{2 + }}}^o - E_{{O_2}/{H_2}O}^o = 1.51 - 1.223 = 0.287\,V$

Since $E_{cell}^o > 0$, therefore net cell reaction is spontaneous and so $MnO_4^ - $ liberate O₂ from H₂O in presence of an acid.

To determine the electromotive force (EMF) of the given cell reaction at 298 K:

Reaction: Ni(s) + 2Ag⁺ (0.001 M) → Ni²⁺ (0.001 M) + 2Ag(s)

Given data:

Standard cell potential, E°_cell = 10.5 V

Value at 298 K, $\frac{2.303\,RT}{F}$ = 0.059

Using the Nernst equation to find the cell potential, E_cell:

$ E_{cell} = E_{cell}^o - \frac{0.059}{n} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} $

Where:

• n = number of moles of electrons transferred = 2


• $[\text{Ni}^{2+}] = 0.001\, \text{M}$


• $[\text{Ag}^+] = 0.001\, \text{M}$

Substituting the values:

$ E_{cell} = 10.5 - \frac{0.059}{2} \log \frac{(10^{-3})}{(10^{-3})^2} $

$ = 10.5 - \frac{0.0295}{2} \log (10^3) $

$ = 10.5 - 0.0295 \times 3 $

$ = 10.5 - 0.0885 $

$ = 10.4115\, \text{V} $

Therefore, the EMF of the cell is 10.4115 V.

Key idea According to Nernst equation,

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303}{n F} R T \log \frac{\left[M_p\right]^{n+}}{\left[M_r\right]^{n+}}$

where, $\left[M_p\right]^{n+}$ and $\left[M_r\right]^{n+}$ are the concentrations of products and reactants respectively.

For the given cell, emf can be calculated as follows:

$E_{\mathrm{emf}}=E_{\mathrm{cell}}^{\circ}-\frac{2.303 R T}{n F} \log \frac{\left[M_p\right]_{(a q)}^{\mathrm{n}+}}{\left[M_r\right]_{(a q)}^{\mathrm{n}+}}$

For the reaction at $25^{\circ} \mathrm{C}$

$\begin{aligned} E_{\text {emf }} & =E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[M_p^{n-1}\right]}{\left[M_r^{n+}\right]} \\ \therefore \quad & E_{\text {cell }}^{\circ}=4 \text { Volt (Given) and } n=2 \\ & E_{\text {emf }}=4-\frac{0.059}{2} \log \frac{10^{-4}}{10^{-2}} \\ & E_{\text {emf }}=4-0.03 \log 10^{-2} \\ E_{\text {emf }} & =4+0.06=4.06 \mathrm{~V} \end{aligned}$

Hence, option (b) is the correct.

Given,

$\begin{aligned} \Delta G_f^{\circ}\left(\mathrm{Ag}_2 \mathrm{O}\right) & =-11.21 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta G_f^{\circ}(\mathrm{ZnO}) & =-318.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

As we know that,

$\Delta G^{\circ}=\Delta G_f^{\circ}(\mathrm{ZnO})-\Delta G_f^{\circ}\left(\mathrm{Ag}_2 \mathrm{O}\right)$

Putting the values in the above given equation, we get

$\begin{array}{lc} \therefore & \Delta G^{\circ}=(-318.30+11.21) \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & =-307.09 \mathrm{~kJ}=-307.09 \times 10^3 \mathrm{~J} \\ \because & \Delta G^{\circ}=-n F E_{\text {cell }}^{\circ} \\ \therefore & -307.09 \times 10^3=-2 \times 96500 \times E_{\text {cell }}^{\circ} \\ \text { or } & E_{\text {cell }}^{\circ}=1.591 \mathrm{~V} \end{array}$

Given,

$\begin{aligned} \lambda_{\mathrm{m}}^{\circ}(\mathrm{NaOH}) & =248 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \\ \lambda_{\mathrm{m}}^{\circ}(\mathrm{NaCl}) & =126 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \\ \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{BaCl}_2\right) & =280 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \end{aligned}$

The reaction for the formation of $\mathrm{Ba}(\mathrm{OH})_2$ can be written as

$\begin{aligned} & \mathrm{BaCl}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{NaCl} \\ & \because \lambda_{\mathrm{m}}^{\circ} \mathrm{Ba}(\mathrm{OH})_2=\lambda_{\mathrm{m}}^{\circ} \mathrm{BaCl}_2+2 \lambda_{\mathrm{m}}^{\circ} \mathrm{NaOH}-2 \lambda_{\mathrm{m}}^{\circ} \mathrm{NaCl} \\ & \therefore \lambda_{\mathrm{m}}^{\circ} \mathrm{Ba}(\mathrm{OH})_2=280 \times 10^{-4}+2 \times 248 \times 10^{-4} \\ &-2 \times 126 \times 10^{-4} \\ &=(280+496-252) \times 10^{-4} \\ &=524 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \end{aligned}$

Both assertion and reason are correct and reason is the correct explanation of assertion. On applying external voltage greater than 1.1V in a Daniell cell, the current flows in the reverse direction i.e. from Zn to Cu (anode to cathode). The electrons flow from copper to zinc. Zn is deposited at Zn electrode and Cu dissolves at Cu electrode. The reaction is

$\mathrm{Zn}^{2+}+\mathrm{Cu} \longrightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}$