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

To determine the ratio \([\text{Fe}^{2+}]/[\text{Fe}^{3+}]\), we use the Nernst equation for the cell potential. The cell's overall reaction is:
\[2\text{Fe}^{3+} + \text{H}_2 \rightarrow 2\text{Fe}^{2+} + 2\text{H}^+.\]
The standard cell potential \(E^\circ\) can be derived from standard reduction potentials:
\[E^\circ = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E^\circ_{\text{H}^+/ \text{H}_2} = 0.771\,\text{V} - 0\,\text{V} = 0.771\,\text{V}.\]
Using the Nernst equation:
\[E = E^\circ - \frac{RT}{nF}\ln Q,\]
where \(E = 0.712\,\text{V}\), \(R = 8.314\,\text{J/mol}\cdot\text{K}\), \(T = 298\,\text{K}\), \(n = 2\), and \(F = 96485\,\text{C/mol}\). The reaction quotient \(Q\) is:
\[Q = \frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2[H^+]^2}.\]
Since \([H^+] = 1\,\text{M}\),
\[Q = \frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}.\]
The Nernst equation becomes:
\[0.712 = 0.771 - \frac{8.314 \times 298}{2 \times 96485}\ln\left(\frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}\right).\]
Solving for \(\ln\left(\frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}\right)\):
\[\ln\left(\frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}\right) = \frac{0.771 - 0.712}{0.0257} = 2.293.\]
\[\frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2} = e^{2.293}.\]
Thus, \(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = e^{1.1465} \approx 3.15\).
Rounding to the nearest integer, the ratio \([\text{Fe}^{2+}]/[\text{Fe}^{3+}]\) is 3.
Verification against range: The calculated value 3 is within the range 10,10, confirming correctness.