Question

Calculate emf of the following cell at 298 K :
\( \text{Sn} \mid \text{Sn}^{2+} (0.001 \text{ M}) \parallel \text{H}^+ (0.01 \text{ M}) \mid \text{H}_{2(g)} (1 \text{ bar}) \mid \text{Pt}_{(s)} \)
Given : \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = - 0.14 \text{ V}, \quad E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \text{ V} \quad [\log 10 = 1] \)

Hint:

Don't forget the stoichiometric coefficient from the balanced equation becomes the power in the reaction quotient \( Q \). Here, \( 2\text{H}^+ \) means \( [\text{H}^+] \) must be squared!

Answer 1

Step 1: Set up the two half-reactions and note n.
Reading the cell notation, tin is oxidised at the anode, $Sn \rightarrow Sn^{2+} + 2e^-$, and hydrogen ion is reduced at the cathode, $2H^+ + 2e^- \rightarrow H_2$, so two electrons pass per formula unit of reaction, giving $n=2$.
Step 2: Find the standard cell potential. \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00 - (-0.14) = 0.14\ \text{V} \]
Step 3: Bring in the Nernst equation for the actual conditions. \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log\frac{[Sn^{2+}]}{[H^+]^2} = 0.14 - \frac{0.0591}{2}\log\frac{10^{-3}}{(10^{-2})^2} \] Since $\dfrac{10^{-3}}{10^{-4}} = 10$ and $\log 10 = 1$, \[ E_{cell} = 0.14 - 0.02955 \times 1 \]
Step 4: Work out the final emf. \[ E_{cell} = 0.14 - 0.02955 = 0.11045\ \text{V} \approx 0.11\ \text{V} \]
The cell therefore delivers a small positive emf under these dilute conditions. \[ \boxed{E_{cell} \approx 0.11\ \text{V}} \]