Q.(i) Why does the conductivity of a solution decrease with dilution? [1]

The cell voltage depends on the reaction quotient via the Nernst equation. As the reaction proceeds, $[Zn^{2+}]$ increases and $[Cu^+]$ decreases, so the voltage decreases. The correct option is (B).

The free energy change of a cell reaction is related to its EMF by Delta G = -nFE, which is the thermodynamic basis for the Nernst equation.

The electrical work done by a galvanic cell is equal to the product of the total charge passed and the EMF of the cell. The total charge passed when n moles of electrons flow is nF (F = Faraday constant = 96500 C/mol).

Maximum electrical work obtainable = nFE (E = EMF of the cell)

This maximum work done by the system equals the decrease in Gibbs free energy of the system, so:

Delta G = -nFE

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At equilibrium a galvanic cell is fully discharged, so its EMF = 0.

As a galvanic cell operates, the concentrations change until the reaction reaches equilibrium (the cell is 'dead').

From the Nernst equation: Ecell = E°cell − (0.059/n) log Q.

At equilibrium Q = K and Ecell = 0, giving the relation E°cell = (0.059/n) log K.

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The standard Gibbs free energy change of a cell reaction is related to the standard cell emf by $\Delta G^\circ = -nFE^\circ_{cell}$.

The maximum electrical work obtainable from a galvanic cell equals the decrease in Gibbs free energy of the cell reaction. The electrical work done is the product of the total charge passed (nF, where n is the number of moles of electrons transferred in the balanced cell reaction and F is Faraday's constant, 96500 C/mol) and the cell's emf:

Electrical work = nFE_cell

Since this work is done at the expense of the free energy of the system, $\Delta G = -nFE_{cell}$, and under standard conditions:

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The standard Gibbs energy change of a reaction and its equilibrium constant are linked by $\Delta G^\circ = -RT\ln K$.

For any reaction at equilibrium, thermodynamics gives the relation

$\Delta G^\circ = -RT\ln K = -2.303\,RT\log K$

where $R$ is the gas constant, $T$ the absolute temperature, and $K$ the equilibrium constant.

This connects directly to electrochemistry through the Nernst equation. Since $\Delta G^\circ = -nFE^\circ_{cell}$ (where $n$ = number of electrons transferred, $F$ = Faraday constant, $E^\circ_{cell}$ = standard cell potential), equating the two expressions for $\Delta G^\circ$ gives

$-nFE^\circ_{cell} = -RT\ln K \quad\Rightarrow\quad E^\circ_{cell} = \frac{RT}{nF}\ln K = \frac{0.0591}{n}\log K \text{ (at 298 K)}$

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For the cell Sn(s) | Sn²⁺ || H⁺ | H₂(g)(1 bar) | Pt(s), n = 2 and the Nernst equation is $E_{cell}=E^\circ_{cell}-\frac{0.0591}{2}\log\frac{[Sn^{2+}]p_{H_2}}{[H^+]^2}$.

From the cell notation, the left electrode (Sn) is the anode (oxidation) and the right electrode (Pt, with H₂/H⁺) is the cathode (reduction):

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

Cathode (reduction): $2H^+(aq) + 2e^- \rightarrow H_2(g)$

Overall cell reaction: $Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)$

Here the number of electrons transferred, $n = 2$.

The general Nernst equation is $E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q$, where $Q$ is the reaction quotient (products over reactants, each raised to its stoichiometric coefficient, gases as partial pressure, solids/pure liquids omitted):

$Q = \dfrac{[Sn^{2+}]\, p_{H_2}}{[H^+]^2}$

So:

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By the Nernst equation, electrode potential depends on the nature of the metal/ions, their concentration and temperature, but not on pressure (for a metal electrode), so option (c).

The potential of an electrode is governed by the Nernst equation:

E = E° - (RT/nF) ln (1/[M^n+])

From this and the nature of the half-cell:

• It depends on the nature of the metal and its ions (through E°).
• It depends on the concentration of the ions in solution (the [M^n+] term). …

E = E° + 0.059·pH for this couple → 1.30 + 0.118 = 1.42 V.

For the quinhydrone-type couple written as H₂Q ⇌ Q + 2H⁺ + 2e⁻ (n = 2), the Nernst equation for the reaction as written is

E = E° − (0.059/2)·log([Q][H⁺]²/[H₂Q]).

Taking activities of quinone and hydroquinone as unity: …

Standard free energy change and standard cell potential are linked by $\Delta G^\circ = -nFE^\circ_{cell}$.

The maximum electrical work obtainable from a galvanic cell equals the decrease in Gibbs free energy of the cell reaction. Electrical work done = charge $\times$ potential = $nFE^\circ_{cell}$, and since this work is done by the system (free energy decreases), we get

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

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