The standard Gibbs energy of formation of aqueous Zn^2+ is -147.1 kJ mol^-1 at 298 K. Taking Delta_f G deg = 0 for Zn(s), H2(g) and H+(aq), and F = 96500 C mol^-1, the standard reduction potential of Zn^2+/Zn is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ec053

The standard Gibbs energy of formation of aqueous $Zn X^{2 +}$ is $- 147.1 kJ mo l^{- 1}$ at 298 K. Taking $Δ_{f} G^{\circ} = 0$ for $Zn (s)$ , $H X_{2} (g)$ and $H X^{+} (aq)$ , and $F = 96500 C mo l^{- 1}$ , the standard reduction potential of $Zn X^{2 +} / Zn$ is

The reduction potential refers to the cell

Pt ∣ H X_{2} ∣ H X^{+} ∣∣ Zn X^{2 +} ∣ Zn

whose reaction is

Zn X^{2 +} + H X_{2} Zn + 2 H X^{+}

Δ_{r} G^{\circ} = 0 - Δ_{f} G^{\circ} (Zn X^{2 +}) = + 147.1 kJ

(reverse of formation). With

n = 2

E^{\circ} = - Δ_{r} G^{\circ} / (n F) = - 147100/ (2 \times 96500) = - 0.762 V \approx - 0.76 V

. The negative sign shows

Zn X^{2 +}

is harder to reduce than

H X^{+}

. Options C and D use

n = 1

and

n = 4