Calculate the cell potential at $298 \ K$ for the following cells:
$(a)$ $Cd \mid Cd^{2+}(0.02 \ M) \parallel H^{+}(1 \ M) \mid H_{2(g)}(1 \ bar) \mid Pt$ $\left[ E_{Cd^{2+} \mid Cd}^0 = -0.40 \ V \right]$
$(b)$ $Al \mid Al^{3+}(0.25 \ M) \parallel Zn^{2+}(0.15 \ M) \mid Zn_{(s)}$ $\left[ E_{Al^{3+} \mid Al}^0 = -1.66 \ V, E_{Zn^{2+} \mid Zn}^0 = -0.76 \ V \right]$

(N/A) The cell reaction is $Cd_{(s)} + 2H^+_{(aq)} \rightarrow Cd^{2+}_{(aq)} + H_{2(g)}$.
$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = 0.00 \ V - (-0.40 \ V) = 0.40 \ V$.
Using the Nernst equation: $E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log \frac{[Cd^{2+}]}{[H^+]^2}$.
$E_{cell} = 0.40 - 0.0295 \log \frac{0.02}{1^2} = 0.40 - 0.0295 \log(2 \times 10^{-2})$.
$E_{cell} = 0.40 - 0.0295 (-1.699) = 0.40 + 0.050 = 0.45 \ V$.
$(b)$ The cell reaction is $2Al_{(s)} + 3Zn^{2+}_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3Zn_{(s)}$.
$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = -0.76 \ V - (-1.66 \ V) = 0.90 \ V$.
Using the Nernst equation with $n=6$: $E_{cell} = E_{cell}^0 - \frac{0.059}{6} \log \frac{[Al^{3+}]^2}{[Zn^{2+}]^3}$.
$E_{cell} = 0.90 - 0.00983 \log \frac{(0.25)^2}{(0.15)^3} = 0.90 - 0.00983 \log \frac{0.0625}{0.003375}$.
$E_{cell} = 0.90 - 0.00983 \log(18.518) = 0.90 - 0.00983(1.2676) = 0.90 - 0.0125 = 0.8875 \ V \approx 0.89 \ V$.