$E^{\circ} (Ag X^{+} / Ag) = 0.80 V$ and $E^{\circ} (Cu X^{2 +} / Cu) = 0.34 V$ . The standard Gibbs energy change (in kJ, two decimals) for $Cu (s) + 2 Ag X^{+} (aq) Cu X^{2 +} (aq) + 2 Ag (s)$ is ( $F = 96500 C mo l^{- 1}$ ).

Numerical answer type, answer in kJ, tolerance 0.02.

Answer: -88.78 kJ

E_{cell}^{\circ} = 0.80 - 0.34 = 0.46 V

and

n = 2

(two

Ag X^{+}

each take one electron from one Cu).

Δ_{r} G^{\circ} = - n F E^{\circ} = - 2 \times 96500 \times 0.46 = - 88780 J = - 88.78 kJ