Question

Consider the cell:
\[\text{Pt(s)}|\text{H}_2(g, 1\,\text{atm})|\text{H}^+(aq, 1\,\text{M})||\text{Fe}^{3+}(aq), \text{Fe}^{2+}(aq)||\text{Pt(s)}.\]
When the potential of the cell is 0.712 V at 298 K, the ratio \([\text{Fe}^{2+}]/[\text{Fe}^{3+}]\) is ____ (Nearest integer).

Hint:

To solve problems involving cell potentials, use the Nernst equation to relate concentrations and the measured potential. Simplify logarithmic terms step-by-step to avoid calculation errors.

Answer 1

Correct Answer: 10

The given cell is:
\[\text{Pt(s)}|\text{H}_2(g, 1\,\text{atm})|\text{H}^+(aq, 1\,\text{M})||\text{Fe}^{3+}(aq), \text{Fe}^{2+}(aq)||\text{Pt(s)}.\]
Step 1: Cell Reactions
At the anode:
\[\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-.\]
At the cathode:
\[\text{Fe}^{3+}_{(aq)} + e^- \rightarrow \text{Fe}^{2+}_{(aq)}.\]
Step 2: Standard Cell Potential
The standard cell potential is:
\[E^\circ = E^\circ_{\text{H}_2/\text{H}^+} + E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}.\]
Substitute the given values:
\[E^\circ = 0 + 0.771 = 0.771\,\text{V}.\]
Step 3: Nernst Equation for the Cell Potential
The cell potential at non-standard conditions is given by the Nernst equation:
\[E = E^\circ - \frac{0.06}{1} \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.\]
Substitute \(E = 0.712\,\text{V}\) and \(E^\circ = 0.771\,\text{V}\):
\[0.712 = 0.771 - 0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.\]
Step 4: Solve for \(\log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\)
Rearrange the equation:
\[0.712 - 0.771 = -0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.\]
\[-0.059 = -0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.\]
\[\log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = \frac{-0.059}{-0.06} = 1.\]
Step 5: Calculate the Ratio
From \(\log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 1\):
\[\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 10^1 = 10.\]
Conclusion: The ratio \(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\) is \(\mathbf{10}\).