For the reaction taking place in the cell: $\mathrm{Pt}(\mathrm{s})|\mathrm{H}_{2}(\mathrm{~g})|\mathrm{H}^{+}(\mathrm{aq}) || \mathrm{Ag}^{+}(\mathrm{aq}) | \mathrm{Ag}(\mathrm{s})$ $\mathrm{E}^{\circ}_{\text{Cell}}=+0.5332 \mathrm{~V}$. The value of $\Delta_{\mathrm{f}} \mathrm{G}^{0}$ is ______ $\mathrm{kJ~mol}^{-1}$. (in nearest integer)

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Redox Reactions and Electrochemistry

For the reaction taking place in the cell:

Pt (s) ∣ H_{2} (g) ∣ H^{+} (aq) ∣∣ Ag^{+} (aq) ∣ Ag (s)

E_{Cell}^{\circ} = + 0.5332 V

.
The value of

Δ_{f} G^{0}

is ______

kJ mol^{- 1}

. (in nearest integer)

Answer: 51

The cell reaction is:

\frac{1}{2} H_{2} + Ag^{+} \to H^{+} + Ag

The relationship between Gibbs free energy and cell potential is:

Δ G^{\circ} = - n F E_{cell}^{\circ}

where

n = 1

(one electron transferred per Ag

^{+}

F = 96500

C mol

^{- 1}

Δ G^{\circ} = - 1 \times 0.5332 \times 96500 = - 51453

= - 51.45

kJ mol

^{- 1}

Rounding to nearest integer:

∣Δ G^{\circ} ∣ = 51

kJ mol

^{- 1}

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer