Question

Consider the cell at \(25^{\circ}C\): \(Zn | Zn^{2+} (aq, 1 \text{ M}) || Fe^{3+} (aq), Fe^{2+} (aq) | Pt(s)\). The fraction of total iron present as \(Fe^{3+}\) ion at the cell potential of 1.500 V is \(x \times 10^{-2}\). The value of \(x\) is ________. (Nearest integer) (Given: \(E^0_{Fe^{3+}/Fe^{2+}} = 0.77 \text{ V}\), \(E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}\))

Hint:

For a 1-electron reduction, a potential difference of 0.059 V corresponds to a tenfold change in the concentration ratio.

Answer 1

Correct Answer: 24

Step 1: Understanding the Concept:
The cell potential is calculated using the Nernst equation, which accounts for non-standard concentrations.
The total iron consists of both \(Fe^{3+}\) and \(Fe^{2+}\) ions.
Step 2: Key Formula or Approach:
1. Standard Cell Potential \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\).
2. Nernst Equation: \(E_{cell} = E^0_{cell} - \frac{0.059}{n} \log \frac{[Products]}{[Reactants]}\).
Step 3: Detailed Explanation:
1. Cell Reaction:
Anode: \(Zn \rightarrow Zn^{2+} + 2e^-\).
Cathode: \(Fe^{3+} + e^- \rightarrow Fe^{2+}\) (multiply by 2 for electron balance).
Net: \(Zn + 2Fe^{3+} \rightarrow Zn^{2+} + 2Fe^{2+}\). (\(n=2\))

2. Calculate \(E^0_{cell}\):
\(E^0_{cell} = 0.77 - (-0.76) = 1.53 \text{ V}\).

3. Apply Nernst Equation:
\[ 1.50 = 1.53 - \frac{0.059}{2} \log \frac{[Zn^{2+}][Fe^{2+}]^2}{[Fe^{3+}]^2} \]
\[ -0.03 = -0.0295 \log \left( \frac{[Fe^{2+}]}{[Fe^{3+}]} \right)^2 \]
\[ \log \frac{[Fe^{2+}]}{[Fe^{3+}]} \approx \frac{0.03}{0.059} \approx 0.508 \]
\[ \frac{[Fe^{2+}]}{[Fe^{3+}]} \approx 10^{0.508} \approx 3.22 \]

4. Calculate Fraction of \(Fe^{3+}\):
Let \([Fe^{3+}] = y\) and \([Fe^{2+}] = 3.22y\).
Total Iron = \(y + 3.22y = 4.22y\).
Fraction \( = \frac{y}{4.22y} = 0.237 \approx 24 \times 10^{-2}\).
Step 4: Final Answer:
The value of \(x\) is 24.