Electrochemistry — NCERT Solutions

Cell Representation:
$$\text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) || \text{Ag}^+(\text{aq}) | \text{Ag(s)}$$

Explanation:
Zn is oxidised (loses electrons) → acts as anode (left).
Ag⁺ is reduced (gains electrons) → acts as cathode (right).

(i) Negatively charged electrode:
The anode (Zn electrode) is negatively charged because it loses electrons, which accumulate on it and flow through the external circuit toward the cathode.

(ii) Carriers of current:

• In the external circuit (metallic wire): Current is carried by electrons flowing from anode (Zn) to cathode (Ag).
• In the electrolytic solution / salt bridge: Current is carried by ions — cations migrate toward the cathode and anions migrate toward the anode.

(iii) Individual electrode reactions:

At Anode (oxidation):
$$\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$$

At Cathode (reduction):
$$2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag(s)}$$

Overall cell reaction:
$$\text{Zn(s)} + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Ag(s)}$$

Standard electrode potentials (from Table):

• $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74\,\text{V}$
• $E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40\,\text{V}$
• $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77\,\text{V}$
• $E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\,\text{V}$

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

Cr is oxidised → anode; Cd²⁺ is reduced → cathode.

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.40 - (-0.74) = +0.34\,\text{V}$$

$n = 6$ electrons transferred.

$$\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -6 \times 96487 \times 0.34$$
$$= -196,672\,\text{J mol}^{-1} \approx -196.67\,\text{kJ mol}^{-1}$$

$$\log K_c = \frac{nE^\circ}{0.0591} = \frac{6 \times 0.34}{0.0591} = \frac{2.04}{0.0591} = 34.52$$

$$K_c = 10^{34.52} \approx 3.31 \times 10^{34}$$

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

Fe²⁺ is oxidised → anode; Ag⁺ is reduced → cathode.

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 - 0.77 = +0.03\,\text{V}$$

$n = 1$ electron transferred.

$$\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -1 \times 96487 \times 0.03$$
$$= -2894.6\,\text{J mol}^{-1} \approx -2.89\,\text{kJ mol}^{-1}$$

$$\log K_c = \frac{nE^\circ}{0.0591} = \frac{1 \times 0.03}{0.0591} = 0.5076$$

$$K_c = 10^{0.5076} \approx 3.22$$

Standard electrode potentials:

• $E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37\,\text{V}$
• $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$
• $E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\,\text{V}$
• $E^\circ_{\text{H}^+/\text{H}_2} = 0.00\,\text{V}$
• $E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14\,\text{V}$
• $E^\circ_{\text{Br}_2/\text{Br}^-} = +1.09\,\text{V}$

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

Cell reaction: $\text{Mg} + \text{Cu}^{2+} \rightarrow \text{Mg}^{2+} + \text{Cu}$; $n = 2$

$E^\circ_{\text{cell}} = 0.34 - (-2.37) = 2.71\,\text{V}$

Nernst equation:
$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2}\log\frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}$$

$$= 2.71 - \frac{0.0591}{2}\log\frac{0.001}{0.0001}$$

$$= 2.71 - 0.02955 \times \log(10)$$

$$= 2.71 - 0.02955 \times 1 = 2.71 - 0.02955$$

$$\boxed{E_{\text{cell}} = 2.68\,\text{V}}$$

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

Cell reaction: $\text{Fe} + 2\text{H}^+ \rightarrow \text{Fe}^{2+} + \text{H}_2$; $n = 2$

$E^\circ_{\text{cell}} = 0.00 - (-0.44) = 0.44\,\text{V}$

Nernst equation:
$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2}\log\frac{[\text{Fe}^{2+}] \cdot p_{\text{H}_2}}{[\text{H}^+]^2}$$

$$= 0.44 - \frac{0.0591}{2}\log\frac{0.001 \times 1}{(1)^2}$$

$$= 0.44 - 0.02955 \times \log(10^{-3})$$

$$= 0.44 - 0.02955 \times (-3)$$

$$= 0.44 + 0.08865$$

$$\boxed{E_{\text{cell}} = 0.529\,\text{V}}$$

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

Cell reaction: $\text{Sn} + 2\text{H}^+ \rightarrow \text{Sn}^{2+} + \text{H}_2$; $n = 2$

$E^\circ_{\text{cell}} = 0.00 - (-0.14) = 0.14\,\text{V}$

Nernst equation:
$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2}\log\frac{[\text{Sn}^{2+}] \cdot p_{\text{H}_2}}{[\text{H}^+]^2}$$

$$= 0.14 - 0.02955 \times \log\frac{0.050 \times 1}{(0.020)^2}$$

$$= 0.14 - 0.02955 \times \log\frac{0.050}{0.0004}$$

$$= 0.14 - 0.02955 \times \log(125)$$

$$= 0.14 - 0.02955 \times 2.097$$

$$= 0.14 - 0.06197$$

$$\boxed{E_{\text{cell}} = 0.078\,\text{V}}$$

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

Here Br⁻ is oxidised at anode: $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$
H⁺ is reduced at cathode: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$

Cell reaction: $2\text{Br}^-(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Br}_2(\text{l}) + \text{H}_2(\text{g})$; $n = 2$

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00 - 1.09 = -1.09\,\text{V}$

Nernst equation:
$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2}\log\frac{p_{\text{H}_2}}{[\text{Br}^-]^2[\text{H}^+]^2}$$

$$= -1.09 - 0.02955 \times \log\frac{1}{(0.010)^2(0.030)^2}$$

$$= -1.09 - 0.02955 \times \log\frac{1}{(10^{-4})(9 \times 10^{-4})}$$

$$= -1.09 - 0.02955 \times \log\frac{1}{9 \times 10^{-8}}$$

$$= -1.09 - 0.02955 \times \log(1.11 \times 10^7)$$

$$= -1.09 - 0.02955 \times 7.046$$

$$= -1.09 - 0.2082$$

$$\boxed{E_{\text{cell}} = -1.298\,\text{V}}$$

(The negative value indicates the reaction is non-spontaneous in this direction.)

Given reaction:
$$\text{Zn(s)} + \text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Ag(s)} + 2\text{OH}^-(\text{aq})$$

Identifying half-reactions:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
$E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$

Cathode (reduction): $\text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} + 2e^- \rightarrow 2\text{Ag(s)} + 2\text{OH}^-(\text{aq})$
$E^\circ_{\text{Ag}_2\text{O}/\text{Ag}} = +0.34\,\text{V}$

(Note: The standard electrode potential for Ag₂O/Ag in alkaline medium is taken as +0.34 V as given in standard tables for this cell.)

Standard cell potential:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = +1.10\,\text{V}$$

Number of electrons transferred: $n = 2$

Standard Gibbs energy:
$$\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -2 \times 96487 \times 1.10$$
$$= -212271.4\,\text{J mol}^{-1}$$

$$\boxed{\Delta_r G^\circ = -212.27\,\text{kJ mol}^{-1}}$$

$$\boxed{E^\circ_{\text{cell}} = 1.10\,\text{V}}$$

Conductivity ($\kappa$):
Conductivity of a solution is defined as the conductance of a solution held between two electrodes of unit cross-sectional area ($A = 1\,\text{m}^2$) and unit distance apart ($l = 1\,\text{m}$).
$$\kappa = G \times \frac{l}{A}$$
SI unit: $\text{S m}^{-1}$ (or $\text{S cm}^{-1}$)

Molar Conductivity ($\Lambda_m$):
Molar conductivity is defined as the conductivity of a solution divided by its molar concentration:
$$\Lambda_m = \frac{\kappa}{c}$$
where $c$ is concentration in mol m⁻³ (or mol L⁻¹). SI unit: $\text{S m}^2\text{mol}^{-1}$ (or $\text{S cm}^2\text{mol}^{-1}$).

Variation with Concentration:

Conductivity ($\kappa$):

• Conductivity decreases with decrease in concentration (dilution).
• On dilution, the number of ions per unit volume decreases, so fewer ions are available to carry charge, reducing conductivity.

Molar Conductivity ($\Lambda_m$):

• Molar conductivity increases with decrease in concentration (dilution).
• For strong electrolytes: $\Lambda_m$ increases slowly and linearly with $\sqrt{c}$ (Debye-Hückel-Onsager equation: $\Lambda_m = \Lambda^\circ_m - A\sqrt{c}$). The increase is due to decreased interionic interactions at lower concentrations.
• For weak electrolytes: $\Lambda_m$ increases steeply at very low concentrations because the degree of dissociation increases significantly on dilution, producing more ions.

Given:

• Concentration, $c = 0.20\,\text{mol L}^{-1} = 0.20 \times 10^{-3}\,\text{mol cm}^{-3}$
• Conductivity, $\kappa = 0.0248\,\text{S cm}^{-1}$

Formula:
$$\Lambda_m = \frac{\kappa}{c}$$

Calculation:
$$\Lambda_m = \frac{0.0248\,\text{S cm}^{-1}}{0.20 \times 10^{-3}\,\text{mol cm}^{-3}}$$

$$= \frac{0.0248}{2 \times 10^{-4}}$$

$$= 124\,\text{S cm}^2\text{mol}^{-1}$$

$$\boxed{\Lambda_m = 124\,\text{S cm}^2\text{mol}^{-1}}$$

Given:

• Resistance, $R = 1500\,\Omega$
• Conductivity, $\kappa = 0.146 \times 10^{-3}\,\text{S cm}^{-1}$

Formula:
Conductance $G = \frac{1}{R}$

Cell constant $= \frac{l}{A} = \kappa \times R = G^*$

$$\text{Cell constant} = \kappa \times R$$

Calculation:
$$\text{Cell constant} = 0.146 \times 10^{-3}\,\text{S cm}^{-1} \times 1500\,\Omega$$

$$= 0.146 \times 10^{-3} \times 1500\,\text{cm}^{-1}$$

$$= 0.219\,\text{cm}^{-1}$$

$$\boxed{\text{Cell constant} = 0.219\,\text{cm}^{-1}}$$

Formula:
$$\Lambda_m = \frac{\kappa}{c}$$

Note: $\kappa$ is given as $10^2 \times \kappa$ in S m⁻¹, so $\kappa = \frac{\text{given value}}{100}\,\text{S m}^{-1}$.

Convert to S cm⁻¹: $1\,\text{S m}^{-1} = 0.01\,\text{S cm}^{-1}$

Concentration in mol cm⁻³: $c\,(\text{mol L}^{-1}) = c \times 10^{-3}\,\text{mol cm}^{-3}$

Calculations:

At c = 0.001 M:
$\kappa = 1.237 \times 10^{-2}\,\text{S m}^{-1} = 1.237 \times 10^{-4}\,\text{S cm}^{-1}$
$$\Lambda_m = \frac{1.237 \times 10^{-4}}{0.001 \times 10^{-3}} = \frac{1.237 \times 10^{-4}}{10^{-6}} = 123.7\,\text{S cm}^2\text{mol}^{-1}$$

At c = 0.010 M:
$\kappa = 11.85 \times 10^{-2}\,\text{S m}^{-1} = 11.85 \times 10^{-4}\,\text{S cm}^{-1}$
$$\Lambda_m = \frac{11.85 \times 10^{-4}}{0.010 \times 10^{-3}} = \frac{11.85 \times 10^{-4}}{10^{-5}} = 118.5\,\text{S cm}^2\text{mol}^{-1}$$

At c = 0.020 M:
$\kappa = 23.15 \times 10^{-4}\,\text{S cm}^{-1}$
$$\Lambda_m = \frac{23.15 \times 10^{-4}}{0.020 \times 10^{-3}} = \frac{23.15 \times 10^{-4}}{2 \times 10^{-5}} = 115.8\,\text{S cm}^2\text{mol}^{-1}$$

At c = 0.050 M:
$\kappa = 55.53 \times 10^{-4}\,\text{S cm}^{-1}$
$$\Lambda_m = \frac{55.53 \times 10^{-4}}{0.050 \times 10^{-3}} = \frac{55.53 \times 10^{-4}}{5 \times 10^{-5}} = 111.1\,\text{S cm}^2\text{mol}^{-1}$$

At c = 0.100 M:
$\kappa = 106.74 \times 10^{-4}\,\text{S cm}^{-1}$
$$\Lambda_m = \frac{106.74 \times 10^{-4}}{0.100 \times 10^{-3}} = \frac{106.74 \times 10^{-4}}{10^{-4}} = 106.74\,\text{S cm}^2\text{mol}^{-1}$$

Summary Table:

Plot: Plot $\Lambda_m$ vs $\sqrt{c}$ — a straight line is obtained. Extrapolate to $\sqrt{c} = 0$ to get $\Lambda^\circ_m$.

Finding $\Lambda^\circ_m$: By extrapolation of the linear plot to zero concentration:
$$\boxed{\Lambda^\circ_m(\text{NaCl}) \approx 126.4\,\text{S cm}^2\text{mol}^{-1}}$$

Given:

• $c = 0.00241\,\text{mol L}^{-1} = 0.00241 \times 10^{-3}\,\text{mol cm}^{-3} = 2.41 \times 10^{-6}\,\text{mol cm}^{-3}$
• $\kappa = 7.896 \times 10^{-5}\,\text{S cm}^{-1}$
• $\Lambda^\circ_m = 390.5\,\text{S cm}^2\text{mol}^{-1}$

Step 1: Calculate molar conductivity
$$\Lambda_m = \frac{\kappa}{c} = \frac{7.896 \times 10^{-5}}{2.41 \times 10^{-6}} = 32.76\,\text{S cm}^2\text{mol}^{-1}$$

$$\boxed{\Lambda_m = 32.76\,\text{S cm}^2\text{mol}^{-1}}$$

Step 2: Calculate degree of dissociation
$$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{32.76}{390.5} = 0.08389 \approx 0.0839$$

Step 3: Calculate dissociation constant

For $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$:
$$K_a = \frac{c\alpha^2}{1-\alpha} = \frac{0.00241 \times (0.0839)^2}{1 - 0.0839}$$

$$= \frac{0.00241 \times 7.039 \times 10^{-3}}{0.9161}$$

$$= \frac{1.696 \times 10^{-5}}{0.9161}$$

$$= 1.851 \times 10^{-5}\,\text{mol L}^{-1}$$

$$\boxed{K_a \approx 1.86 \times 10^{-5}\,\text{mol L}^{-1}}$$

Formula: $Q = n \times F$, where $n$ = number of moles of electrons, $F = 96487\,\text{C mol}^{-1}$

(i) Al³⁺ + 3e⁻ → Al

3 moles of electrons required per mole of Al³⁺.
$$Q = 3 \times 96487 = 289461\,\text{C} \approx 2.89 \times 10^5\,\text{C}$$

$$\boxed{Q = 2.89 \times 10^5\,\text{C}}$$

(ii) Cu²⁺ + 2e⁻ → Cu

2 moles of electrons required per mole of Cu²⁺.
$$Q = 2 \times 96487 = 192974\,\text{C} \approx 1.93 \times 10^5\,\text{C}$$

$$\boxed{Q = 1.93 \times 10^5\,\text{C}}$$

(iii) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

5 moles of electrons required per mole of MnO₄⁻.
$$Q = 5 \times 96487 = 482435\,\text{C} \approx 4.82 \times 10^5\,\text{C}$$

$$\boxed{Q = 4.82 \times 10^5\,\text{C}}$$

(i) Production of Ca from molten CaCl₂

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

Molar mass of Ca = 40 g mol⁻¹

Moles of Ca = $\frac{20.0}{40} = 0.5\,\text{mol}$

Each mole of Ca requires 2 Faradays.

$$\text{Electricity required} = 0.5 \times 2 = 1\,\text{F}$$

$$\boxed{1\,\text{Faraday}}$$

(ii) Production of Al from molten Al₂O₃

Electrode reaction: $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$

Molar mass of Al = 27 g mol⁻¹

Moles of Al = $\frac{40.0}{27} = 1.481\,\text{mol}$

Each mole of Al requires 3 Faradays.

$$\text{Electricity required} = 1.481 \times 3 = 4.44\,\text{F}$$

$$\boxed{4.44\,\text{Faradays}}$$

(i) Oxidation of 1 mol of H₂O to O₂

Half-reaction: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$

For 2 mol H₂O → 4 electrons are transferred.
For 1 mol H₂O → 2 electrons are transferred.

$$Q = 2 \times 96487 = 192974\,\text{C}$$

$$\boxed{Q \approx 1.93 \times 10^5\,\text{C}}$$

(ii) Oxidation of 1 mol of FeO to Fe₂O₃

In FeO: Fe is in +2 state. In Fe₂O₃: Fe is in +3 state.

Half-reaction: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

For 1 mol of FeO (containing 1 mol Fe²⁺), 1 electron is transferred.

$$Q = 1 \times 96487 = 96487\,\text{C}$$

$$\boxed{Q \approx 96487\,\text{C} \approx 9.65 \times 10^4\,\text{C}}$$

Given:

• Current, $I = 5\,\text{A}$
• Time, $t = 20\,\text{min} = 20 \times 60 = 1200\,\text{s}$
• Molar mass of Ni = 58.7 g mol⁻¹
• Electrode reaction: $\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$ ($n = 2$)

Step 1: Calculate charge passed
$$Q = I \times t = 5 \times 1200 = 6000\,\text{C}$$

Step 2: Calculate moles of electrons
$$\text{Moles of electrons} = \frac{Q}{F} = \frac{6000}{96487} = 0.06217\,\text{mol}$$

Step 3: Calculate moles of Ni deposited
$$\text{Moles of Ni} = \frac{\text{moles of electrons}}{2} = \frac{0.06217}{2} = 0.03108\,\text{mol}$$

Step 4: Calculate mass of Ni
$$m = 0.03108 \times 58.7 = 1.825\,\text{g}$$

$$\boxed{m(\text{Ni}) \approx 1.825\,\text{g}}$$

Given:

• Current, $I = 1.5\,\text{A}$
• Mass of Ag deposited = 1.45 g
• Molar mass: Ag = 108 g mol⁻¹, Cu = 63.5 g mol⁻¹, Zn = 65.4 g mol⁻¹
• Electrode reactions: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ ($n=1$); $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ ($n=2$); $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$ ($n=2$)

Step 1: Find moles of Ag deposited
$$\text{Moles of Ag} = \frac{1.45}{108} = 0.01343\,\text{mol}$$

Step 2: Find charge passed (from Ag)
Since $n = 1$ for Ag:
$$Q = 0.01343 \times 96487 = 1295.6\,\text{C}$$

Step 3: Find time
$$t = \frac{Q}{I} = \frac{1295.6}{1.5} = 863.7\,\text{s} \approx 864\,\text{s}$$

$$\boxed{t \approx 864\,\text{s} \approx 14.4\,\text{minutes}}$$

Step 4: Mass of Cu deposited (Cell C)
$$\text{Moles of Cu} = \frac{Q}{2F} = \frac{1295.6}{2 \times 96487} = 6.713 \times 10^{-3}\,\text{mol}$$
$$m(\text{Cu}) = 6.713 \times 10^{-3} \times 63.5 = 0.4263\,\text{g}$$

$$\boxed{m(\text{Cu}) \approx 0.426\,\text{g}}$$

Step 5: Mass of Zn deposited (Cell A)
$$\text{Moles of Zn} = \frac{Q}{2F} = 6.713 \times 10^{-3}\,\text{mol}$$
$$m(\text{Zn}) = 6.713 \times 10^{-3} \times 65.4 = 0.4390\,\text{g}$$

$$\boxed{m(\text{Zn}) \approx 0.439\,\text{g}}$$

Concept: A reaction is feasible (spontaneous) if $E^\circ_{\text{cell}} > 0$.

Standard electrode potentials:

• $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77\,\text{V}$
• $E^\circ_{\text{I}_2/\text{I}^-} = +0.54\,\text{V}$
• $E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\,\text{V}$
• $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$
• $E^\circ_{\text{Br}_2/\text{Br}^-} = +1.09\,\text{V}$

(i) Fe³⁺(aq) + I⁻(aq)

Possible reaction: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.77 - 0.54 = +0.23\,\text{V}$

$E^\circ_{\text{cell}} > 0$ → Reaction is feasible. ✓

(ii) Ag⁺(aq) and Cu(s)

Possible reaction: $2\text{Ag}^+ + \text{Cu} \rightarrow 2\text{Ag} + \text{Cu}^{2+}$

$E^\circ_{\text{cell}} = 0.80 - 0.34 = +0.46\,\text{V}$

$E^\circ_{\text{cell}} > 0$ → Reaction is feasible. ✓

(iii) Fe³⁺(aq) and Br⁻(aq)

Possible reaction: $2\text{Fe}^{3+} + 2\text{Br}^- \rightarrow 2\text{Fe}^{2+} + \text{Br}_2$

$E^\circ_{\text{cell}} = 0.77 - 1.09 = -0.32\,\text{V}$

$E^\circ_{\text{cell}} < 0$ → Reaction is not feasible. ✗

(iv) Ag(s) and Fe³⁺(aq)

Possible reaction: $\text{Ag} + \text{Fe}^{3+} \rightarrow \text{Ag}^+ + \text{Fe}^{2+}$

$E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E^\circ_{\text{Ag}^+/\text{Ag}} = 0.77 - 0.80 = -0.03\,\text{V}$

$E^\circ_{\text{cell}} < 0$ → Reaction is not feasible. ✗

(v) Br₂(aq) and Fe²⁺(aq)

Possible reaction: $\text{Br}_2 + 2\text{Fe}^{2+} \rightarrow 2\text{Br}^- + 2\text{Fe}^{3+}$

$E^\circ_{\text{cell}} = E^\circ_{\text{Br}_2/\text{Br}^-} - E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 1.09 - 0.77 = +0.32\,\text{V}$

$E^\circ_{\text{cell}} > 0$ → Reaction is feasible. ✓

(i) Aqueous AgNO₃ with silver electrodes:

At Cathode: Ag⁺ ions are preferentially discharged (higher reduction potential than H₂O):
$$\text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag(s)}$$
Silver is deposited at the cathode.

At Anode: The silver anode dissolves (since silver is an active electrode, it is easier to oxidise Ag than water/NO₃⁻):
$$\text{Ag(s)} \rightarrow \text{Ag}^+(\text{aq}) + e^-$$
Silver dissolves from the anode.

This is the basis of electroplating and electrolytic refining of silver.

(ii) Aqueous AgNO₃ with platinum electrodes:

At Cathode: Ag⁺ ions are preferentially reduced:
$$\text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag(s)}$$
Silver is deposited.

At Anode: Pt is an inert electrode. Water is oxidised (NO₃⁻ is not easily oxidised):
$$2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^-$$
Oxygen gas is evolved at the anode.

(iii) Dilute H₂SO₄ with platinum electrodes:

At Cathode: H⁺ ions are reduced:
$$2\text{H}^+(\text{aq}) + 2e^- \rightarrow \text{H}_2(\text{g})$$
Hydrogen gas is evolved.

At Anode: Water is oxidised (in dilute H₂SO₄, SO₄²⁻ is not discharged):
$$2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^-$$
Oxygen gas is evolved.

Overall: Water is decomposed into H₂ and O₂.

(iv) Aqueous CuCl₂ with platinum electrodes:

At Cathode: Cu²⁺ ions are preferentially reduced (higher reduction potential than H₂O):
$$\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$$
Copper is deposited.