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: Concept
Nernst equation for non-standard cell potential.

Step 2: Meaning
$E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q$.

Step 3: Analysis
Reaction: $Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)$. Here, $n = 2$.
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00~V - (-0.14~V) = +0.14~V$.
$Q = \frac{[Sn^{2+}][P_{H_2}]}{[H^+]^2} = \frac{(0.001)(1)}{(0.01)^2} = \frac{10^{-3}}{10^{-4}} = 10$.
$E_{cell} = 0.14 - \frac{0.0591}{2} \log(10) = 0.14 - 0.0295(1)$.

Step 4: Conclusion
$E_{cell} = 0.14 - 0.0295 = 0.1105~V$.

Final Answer: $0.1105~V$