A student builds the galvanic cell shown in the figure…

(a) Standard emf. Zinc has the lower potential, so it is the anode. $E^\circ_{\text{cell}} = 0.80 - (-0.76)\ \text{V} = 1.56\ \text{V}$.

(b) Emf at the start. $Q = \dfrac{0.0500}{(0.0200)^2} = 125$, and $\log Q = 2.0969$.

The two electrode potentials give the same: $E_{\text{Ag}} = 0.80 + 0.059\log 0.0200 = 0.6998\ \text{V}$ and $E_{\text{Zn}} = -0.76 + 0.0295\log 0.0500 = -0.7984\ \text{V}$.

(c) Gibbs energy. $\Delta_r G = -nFE = -2 \times 96487 \times 1.498\ \text{J mol}^{-1} = -289.1\ \text{kJ mol}^{-1}$.

(d) Equilibrium constant. $\log K = \dfrac{nE^\circ}{0.059} = \dfrac{2 \times 1.56}{0.059} = 52.881$, so $K = 7.6 \times 10^{52}$. The reaction goes to completion for all practical purposes.

(e) The cell runs down. The emf falls by $0.16\ \text{V}$, so $\log Q = \dfrac{0.16}{0.0295} = 5.4237$ and $Q = 2.653 \times 10^{5}$. If $[\text{Ag}^+]$ falls to $x$, then $[\text{Zn}^{2+}] = 0.0500 + \dfrac{0.0200 - x}{2}$. Solve $\dfrac{[\text{Zn}^{2+}]}{x^2} = 2.653 \times 10^{5}$ by trial: $x = 4.746 \times 10^{-4}\ \text{mol L}^{-1}$, with $[\text{Zn}^{2+}] = 0.0598\ \text{mol L}^{-1}$. Keeping $[\text{Zn}^{2+}]$ at its first value would give $4.34 \times 10^{-4}\ \text{mol L}^{-1}$.