Standard Gibbs energies of formation at 298 K: $Δ_{f} G^{\circ} (Zn X^{2 +}, a q) = - 147.1 kJ mo l^{- 1}$ and $Δ_{f} G^{\circ} (Ag X^{+}, a q) = + 77.1 kJ mo l^{- 1}$ (elements in their standard states have $Δ_{f} G^{\circ} = 0$ ). The standard EMF (in V, two decimals) of the cell $Zn ∣ Zn X^{2 +} (1 M) ∣∣ Ag X^{+} (1 M) ∣ Ag$ is ( $F = 96500 C mo l^{- 1}$ ).

Numerical answer type, answer in V, tolerance 0.01.

Answer: 1.56 V

Cell reaction:

Zn + 2 Ag X^{+} Zn X^{2 +} + 2 Ag

n = 2

Δ_{r} G^{\circ} = Δ_{f} G^{\circ} (Zn X^{2 +}) - 2 Δ_{f} G^{\circ} (Ag X^{+}) = - 147.1 - 2 (77.1) = - 301.3 kJ

E^{\circ} = - Δ_{r} G^{\circ} / (n F) = 301300/ (2 \times 96500) = 1.561 V \approx 1.56 V

. This agrees with

0.80 - (- 0.76) = 1.56 V

from tabulated potentials.