Question

Calculate emf of the following cell at 298 K:
$Sn | Sn^{2+}(0.001~M) || H^+(0.01~M) | H_{2(g)}(1~bar) | Pt_{(s)}$
Given: $E^\circ_{Sn^{2+}/Sn} = -0.14~V$, $E^\circ_{H^+/H_2} = 0.00~V [\log 10 = 1]$

Hint:

Remember to square the $[H^+]$ concentration in the reaction quotient $Q$ due to the stoichiometry.

Answer 1

Step 1: Cell reaction and n.
The overall reaction is $Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)$. Tin loses 2 electrons, so $n = 2$.
Step 2: Calculate $E^\circ_{cell}$.
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00 - (-0.14) = +0.14~V$.
Step 3: Calculate reaction quotient Q.
\[ Q = \frac{[Sn^{2+}] \cdot P_{H_2}}{[H^+]^2} = \frac{0.001 \times 1}{(0.01)^2} = \frac{10^{-3}}{10^{-4}} = 10 \]
Step 4: Apply Nernst equation.
\[ E_{cell} = 0.14 - \frac{0.0591}{2} \log(10) = 0.14 - \frac{0.0591}{2} \times 1 = 0.14 - 0.02955 \approx 0.1105~V \] The EMF of the cell at 298 K is $0.1105~V$.