Q.What is the relation between standard Gibbs' free energy and standard emf of the cell?

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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