Given E deg(Ce^4+/Ce^3+) = 1.61 V and E deg(Zn^2+/Zn) = -0.76 V. Delta_r G deg for Zn + 2Ce^4+ -> Zn^2+ + 2Ce^3+ is (F = 96500 C mol^-1)

JEE Main levelSingle correct optionDifficulty 2 of 3ec049

Given $E^{\circ} (Ce X^{4 +} / Ce X^{3 +}) = 1.61 V$ and $E^{\circ} (Zn X^{2 +} / Zn) = - 0.76 V$ . $Δ_{r} G^{\circ}$ for $Zn + 2 Ce X^{4 +} Zn X^{2 +} + 2 Ce X^{3 +}$ is ( $F = 96500 C mo l^{- 1}$ )

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 1.61 - (- 0.76) = 2.37 V

. Two electrons are transferred per Zn (the Ce half reaction is multiplied by 2 but its potential is not).

Δ_{r} G^{\circ} = - 2 \times 96500 \times 2.37 = - 457410 J = - 457.41 kJ mo l^{- 1}

. Option B uses

n = 1

; option C uses

1.61 - 0.76 = 0.85 V

, a sign error.