Electrochemistry

Given/Concept: The standard electrode potential is always measured relative to the Standard Hydrogen Electrode (SHE), whose potential is taken as zero.

Method:
1. Set up a galvanic cell by connecting the Mg²⁺|Mg half-cell with the Standard Hydrogen Electrode (SHE).
- The cell is: $\text{Mg(s)} | \text{Mg}^{2+}(1\,\text{M}) || \text{H}^+(1\,\text{M}) | \text{H}_2(1\,\text{bar}) | \text{Pt(s)}$
2. Maintain all species at unit activity (1 M concentration for ions, 1 bar pressure for gases, 298 K).
3. Measure the EMF of the cell using a voltmeter.
4. Since Mg is a stronger reducing agent than H₂, Mg acts as the anode and SHE acts as the cathode.
5. The measured cell potential gives:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
$$E^\circ_{\text{cell}} = E^\circ_{\text{SHE}} - E^\circ_{\text{Mg}^{2+}/\text{Mg}}$$
$$E^\circ_{\text{cell}} = 0 - E^\circ_{\text{Mg}^{2+}/\text{Mg}}$$

The experimentally measured value is $E^\circ_{\text{cell}} = +2.37\,\text{V}$, so $E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37\,\text{V}$.

Given: Standard electrode potentials:
- $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$
- $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$

Concept: A spontaneous reaction occurs when the cell EMF is positive, i.e., the metal with lower (more negative) electrode potential displaces the metal with higher electrode potential from its salt solution.

Working:
If copper sulphate is stored in a zinc pot, the following redox reaction would occur:
$$\text{Zn(s)} + \text{CuSO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{Cu(s)}$$

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

Since E^\circ_{\text{cell}} > 0, the reaction is spontaneous. Zinc will dissolve and copper will be deposited.

Conclusion: No, copper sulphate solution cannot be stored in a zinc pot because zinc is more reactive than copper and will displace copper from the copper sulphate solution, corroding the zinc pot.

Concept: A substance can oxidise Fe²⁺ to Fe³⁺ if its standard electrode potential is greater than that of the Fe³⁺/Fe²⁺ couple.

Given: $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77\,\text{V}$

Any oxidising agent (species on the left side of a half-reaction) with E^\circ > +0.77\,\text{V} can oxidise Fe²⁺ to Fe³⁺.

Three such substances:

1. Fluorine (F₂): E^\circ_{\text{F}_2/\text{F}^-} = +2.87\,\text{V} > +0.77\,\text{V} ✓
2. Chlorine (Cl₂): E^\circ_{\text{Cl}_2/\text{Cl}^-} = +1.36\,\text{V} > +0.77\,\text{V} ✓
3. Acidified permanganate (MnO₄⁻/Mn²⁺): E^\circ = +1.51\,\text{V} > +0.77\,\text{V} ✓

(Other acceptable answers include $\text{Br}_2$, $\text{Cr}_2\text{O}_7^{2-}$, $\text{H}_2\text{O}_2$, etc., all having E^\circ > +0.77\,\text{V}.)

Given: pH = 10, so $[\text{H}^+] = 10^{-10}\,\text{M}$

Half-cell reaction:
$$\text{H}^+(\text{aq}) + e^- \rightarrow \frac{1}{2}\text{H}_2(\text{g})$$

Nernst equation for hydrogen electrode:
$$E_{\text{H}^+/\text{H}_2} = E^\circ_{\text{H}^+/\text{H}_2} - \frac{0.0591}{1}\log\frac{1}{[\text{H}^+]}$$

(At 298 K, $\frac{RT}{F}\ln = \frac{0.0591}{n}\log$; $n = 1$; $E^\circ = 0\,\text{V}$; $p_{\text{H}_2} = 1\,\text{bar}$)

$$E = 0 - 0.0591 \times \log\frac{1}{10^{-10}}$$

$$E = -0.0591 \times \log(10^{10})$$

$$E = -0.0591 \times 10$$

$$\boxed{E = -0.591\,\text{V}}$$

Given:
- $[\text{Ag}^+] = 0.002\,\text{M}$
- $[\text{Ni}^{2+}] = 0.160\,\text{M}$
- $E^\circ_{\text{cell}} = 1.05\,\text{V}$
- $n = 2$ (2 electrons transferred)
- Temperature = 298 K

Nernst Equation:
$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q$$

Reaction quotient Q:
$$Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} = \frac{0.160}{(0.002)^2} = \frac{0.160}{4 \times 10^{-6}} = 4 \times 10^4$$

Substituting:
$$E_{\text{cell}} = 1.05 - \frac{0.0591}{2}\log(4 \times 10^4)$$

$$= 1.05 - 0.02955 \times \log(4 \times 10^4)$$

$$\log(4 \times 10^4) = \log 4 + 4 = 0.602 + 4 = 4.602$$

$$E_{\text{cell}} = 1.05 - 0.02955 \times 4.602$$

$$= 1.05 - 0.1360$$

$$\boxed{E_{\text{cell}} \approx 0.914\,\text{V}}$$

*(The textbook answer is given as 0.91 V.)*

Given:
- $E^\circ_{\text{cell}} = 0.236\,\text{V}$
- $n = 2$ (2 electrons transferred)
- $T = 298\,\text{K}$
- $F = 96487\,\text{C mol}^{-1}$
- $R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}$

Step 1: Calculate $\Delta_r G^\circ$
$$\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$$
$$= -2 \times 96487 \times 0.236$$
$$= -45541.8\,\text{J mol}^{-1}$$
$$\boxed{\Delta_r G^\circ \approx -45.54\,\text{kJ mol}^{-1}}$$

Step 2: Calculate equilibrium constant $K_c$

Using $\Delta_r G^\circ = -RT\ln K_c$:
$$\ln K_c = \frac{-\Delta_r G^\circ}{RT} = \frac{45541.8}{8.314 \times 298} = \frac{45541.8}{2477.6} = 18.38$$

$$K_c = e^{18.38}$$

Alternatively, using:
$$\log K_c = \frac{nE^\circ_{\text{cell}}}{0.0591} = \frac{2 \times 0.236}{0.0591} = \frac{0.472}{0.0591} = 7.987$$

$$K_c = 10^{7.987} \approx 9.62 \times 10^7$$

$$\boxed{K_c \approx 9.62 \times 10^7}$$

Concept: Conductivity ($\kappa$) is defined as the conductance of a solution held between electrodes of unit area and unit distance apart. It depends on the number of ions per unit volume of the solution.

Explanation:
Conductivity of an electrolytic solution depends on the number of ions present per unit volume. On dilution, the concentration of the solution decreases, which means the number of ions per unit volume (per cm³ or per m³) decreases. Since fewer ions are available to carry charge per unit volume, the conductivity decreases with dilution.

Conclusion: As dilution increases → concentration decreases → fewer ions per unit volume → conductivity ($\kappa$) decreases.

Concept: Water is a weak electrolyte and its $\Lambda^\circ_m$ cannot be determined by direct extrapolation of $\Lambda_m$ vs $\sqrt{c}$ plot (as done for strong electrolytes). We use Kohlrausch's Law of Independent Migration of Ions.

Method:
Water dissociates as: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$

Using Kohlrausch's law:
$$\Lambda^\circ_m(\text{H}_2\text{O}) = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{OH}^-)$$

This can be obtained from the $\Lambda^\circ_m$ values of strong electrolytes:
$$\Lambda^\circ_m(\text{H}_2\text{O}) = \Lambda^\circ_m(\text{HCl}) + \Lambda^\circ_m(\text{NaOH}) - \Lambda^\circ_m(\text{NaCl})$$

All three of these are strong electrolytes whose $\Lambda^\circ_m$ values can be determined by extrapolation of their $\Lambda_m$ vs $\sqrt{c}$ plots to zero concentration.

Conclusion: $\Lambda^\circ_m$ of water is determined indirectly using Kohlrausch's law by combining the $\Lambda^\circ_m$ values of HCl, NaOH, and NaCl.

Given:
- $c = 0.025\,\text{mol L}^{-1}$
- $\Lambda_m = 46.1\,\text{S cm}^2\text{mol}^{-1}$
- $\lambda^\circ(\text{H}^+) = 349.6\,\text{S cm}^2\text{mol}^{-1}$
- $\lambda^\circ(\text{HCOO}^-) = 54.6\,\text{S cm}^2\text{mol}^{-1}$

Step 1: Calculate $\Lambda^\circ_m$ for methanoic acid (HCOOH)
$$\Lambda^\circ_m(\text{HCOOH}) = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{HCOO}^-)$$
$$= 349.6 + 54.6 = 404.2\,\text{S cm}^2\text{mol}^{-1}$$

Step 2: Calculate degree of dissociation ($\alpha$)
$$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{46.1}{404.2} = 0.114$$

$$\boxed{\alpha = 0.114}$$

Step 3: Calculate dissociation constant ($K_a$)

For $\text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^-$:
$$K_a = \frac{c\alpha^2}{1-\alpha}$$
$$= \frac{0.025 \times (0.114)^2}{1 - 0.114}$$
$$= \frac{0.025 \times 0.012996}{0.886}$$
$$= \frac{3.249 \times 10^{-4}}{0.886}$$
$$= 3.67 \times 10^{-4}\,\text{mol L}^{-1}$$

$$\boxed{K_a = 3.67 \times 10^{-4}\,\text{mol L}^{-1}}$$

Given:
- Current, $I = 0.5\,\text{A}$
- Time, $t = 2\,\text{hours} = 2 \times 3600 = 7200\,\text{s}$
- Faraday constant, $F = 96487\,\text{C mol}^{-1}$
- Charge on one electron, $e = 1.6 \times 10^{-19}\,\text{C}$

Step 1: Calculate total charge (Q)
$$Q = I \times t = 0.5 \times 7200 = 3600\,\text{C}$$

Step 2: Calculate number of electrons
$$\text{Number of electrons} = \frac{Q}{e} = \frac{3600}{1.6 \times 10^{-19}}$$

$$= 2.25 \times 10^{22}\,\text{electrons}$$

$$\boxed{n_e = 2.25 \times 10^{22}\,\text{electrons}}$$

Concept: Metals that cannot be reduced by chemical reducing agents (because they are highly reactive) are extracted by electrolysis of their molten salts or oxides.

List of metals extracted electrolytically:

1. Sodium (Na) — by electrolysis of molten NaCl (Down's process)
2. Magnesium (Mg) — by electrolysis of molten MgCl₂
3. Aluminium (Al) — by electrolysis of molten Al₂O₃ dissolved in cryolite (Hall-Héroult process)
4. Calcium (Ca) — by electrolysis of molten CaCl₂
5. Potassium (K) — by electrolysis of molten KCl
6. Copper (Cu) — electrolytic refining of impure copper

These are all metals with very negative electrode potentials (highly reactive), making chemical reduction impractical.

Given:
- Reaction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
- 1 mol of $\text{Cr}_2\text{O}_7^{2-}$ requires 6 moles of electrons
- $F = 96487\,\text{C mol}^{-1}$

Calculation:
$$Q = n \times F = 6 \times 96487$$
$$= 578922\,\text{C}$$

$$\boxed{Q \approx 5.79 \times 10^5\,\text{C}}$$

During discharging (normal use), the reactions are:
- Anode: $\text{Pb(s)} + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{PbSO}_4(\text{s}) + 2e^-$
- Cathode: $\text{PbO}_2(\text{s}) + \text{SO}_4^{2-}(\text{aq}) + 4\text{H}^+(\text{aq}) + 2e^- \rightarrow \text{PbSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l})$

During recharging, an external electrical energy source reverses the electrode reactions:

- At anode (now acts as cathode during recharge):
$$\text{PbSO}_4(\text{s}) + 2e^- \rightarrow \text{Pb(s)} + \text{SO}_4^{2-}(\text{aq})$$

- At cathode (now acts as anode during recharge):
$$\text{PbSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{PbO}_2(\text{s}) + \text{SO}_4^{2-}(\text{aq}) + 4\text{H}^+(\text{aq}) + 2e^-$$

Overall recharging reaction:
$$2\text{PbSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \xrightarrow{\text{electrical energy}} \text{Pb(s)} + \text{PbO}_2(\text{s}) + 2\text{H}_2\text{SO}_4(\text{aq})$$

Materials involved: PbSO₄ (at both electrodes), H₂O, Pb (reformed at one electrode), PbO₂ (reformed at other electrode), and H₂SO₄ (electrolyte regenerated).

Concept: A fuel cell converts chemical energy of a fuel directly into electrical energy through electrochemical reactions. The fuel must be oxidisable and produce a clean reaction.

Two materials (other than hydrogen) that can be used as fuels in fuel cells:

1. Methane (CH₄): Natural gas can be used in solid oxide fuel cells. It undergoes oxidation at the anode to produce CO₂ and H₂O, releasing electrons.

2. Methanol (CH₃OH): Used in Direct Methanol Fuel Cells (DMFC). Methanol is oxidised at the anode:
$$\text{CH}_3\text{OH} + \text{H}_2\text{O} \rightarrow \text{CO}_2 + 6\text{H}^+ + 6e^-$$

Other acceptable answers include ethanol, propane, etc.

Concept: Rusting of iron is an electrochemical process involving oxidation and reduction reactions occurring at different sites on the iron surface, effectively setting up a galvanic cell.

Mechanism:

When iron is exposed to moist air (water containing dissolved CO₂ or O₂), different parts of the iron surface act as anode and cathode:

At the Anode (oxidation — iron dissolves):
$$\text{Fe(s)} \rightarrow \text{Fe}^{2+}(\text{aq}) + 2e^-$$

At the Cathode (reduction — oxygen is reduced):
$$\text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^- \rightarrow 2\text{H}_2\text{O}(\text{l})$$
(In neutral/slightly acidic water with dissolved CO₂ providing H⁺)

Overall: The Fe²⁺ ions formed at the anode are further oxidised by atmospheric oxygen:
$$4\text{Fe}^{2+} + \text{O}_2 + 4\text{H}_2\text{O} \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{H}^+$$

$\text{Fe}_2\text{O}_3$ combines with water to form hydrated iron(III) oxide, i.e., rust ($\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$).

The electrons flow through the iron from anode to cathode (like in a galvanic cell), and ions migrate through the moisture film (electrolyte). This is exactly analogous to a galvanic cell.

Concept: A metal with a lower (more negative) standard electrode potential is a stronger reducing agent and can displace metals with higher electrode potentials from their salt solutions.

Standard electrode potentials (from Table):
- $E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37\,\text{V}$
- $E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66\,\text{V}$
- $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$
- $E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\,\text{V}$
- $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$

Order of increasing electrode potential (increasing oxidising power / decreasing reducing power):
\text{Mg} < \text{Al} < \text{Zn} < \text{Fe} < \text{Cu}

Order in which they displace each other (decreasing reactivity / reducing power):
\boxed{\text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Cu}}

Mg displaces all others; Al displaces Zn, Fe, Cu; Zn displaces Fe and Cu; Fe displaces Cu; Cu cannot displace any of the above.

Concept: Reducing power of a metal is inversely related to its standard electrode potential. A metal with a more negative $E^\circ$ is a stronger reducing agent (greater tendency to lose electrons).

Given standard electrode potentials:
| Metal | $E^\circ$ (V) |
|-------|---------------|
| K | −2.93 |
| Mg | −2.37 |
| Cr | −0.74 |
| Hg | +0.79 |
| Ag | +0.80 |

Reducing power increases as $E^\circ$ decreases (becomes more negative).

Increasing order of reducing power:
\boxed{\text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K}}

Ag has the highest electrode potential → weakest reducing agent; K has the lowest electrode potential → strongest reducing agent.

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:
| c (M) | $\sqrt{c}$ (M^½) | $\Lambda_m$ (S cm² mol⁻¹) |
|--------|-----------------|---------------------------|
| 0.001 | 0.0316 | 123.7 |
| 0.010 | 0.1000 | 118.5 |
| 0.020 | 0.1414 | 115.8 |
| 0.050 | 0.2236 | 111.1 |
| 0.100 | 0.3162 | 106.74 |

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

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(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. ✓

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(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. ✓

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(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. ✗

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(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. ✗

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(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.

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(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.

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(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.

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(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.

At Anode: Cl⁻ ions are preferentially oxidised (overpotential considerations favour Cl⁻ over H₂O in concentrated solution):
$$2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2e^-$$
Chlorine gas is evolved at the anode.