For mercury, E deg(Hg2^2+/Hg) = 0.80 V for Hg2^2+ + 2e- -> 2Hg and E deg(Hg^2+/Hg2^2+) = 0.92 V for 2Hg^2+ + 2e- -> Hg2^2+. The standard Gibbs energy change (in kJ per mole of Hg^2+, two decimals) for Hg^2+ + 2e- -> Hg is (F = 96500 C mol^-1).

JEE Advanced levelNumerical answerDifficulty 3 of 3ec075

For mercury, $E^{\circ} (Hg X_{2} X^{2 +} / Hg) = 0.80 V$ for $Hg X_{2} X^{2 +} + 2 e X^{-} 2 Hg$ and $E^{\circ} (Hg X^{2 +} / Hg X_{2} X^{2 +}) = 0.92 V$ for $2 Hg X^{2 +} + 2 e X^{-} Hg X_{2} X^{2 +}$ . The standard Gibbs energy change (in kJ per mole of $Hg X^{2 +}$ , two decimals) for $Hg X^{2 +} + 2 e X^{-} Hg$ is ( $F = 96500 C mo l^{- 1}$ ).

Numerical answer type, answer in kJ, tolerance 0.02.

Add the two half reactions as written:

2 Hg X^{2 +} + 4 e X^{-} 2 Hg

with

Δ G^{\circ} = - 2 F (0.92) - 2 F (0.80) = - 2 F (1.72)

. Per mole of

Hg X^{2 +}

(2 electrons):

Δ G^{\circ} = - F (1.72) = - 96500 \times 1.72 = - 165980 J = - 165.98 kJ

, which corresponds to

E^{\circ} (Hg X^{2 +} / Hg) = 1.72/2 = 0.86 V

. Note that

0.92 V

refers to one electron per

Hg X^{2 +}

and

0.80 V

to one electron per Hg atom, so each contributes equally per mole of

Hg X^{2 +}