Question

What would be the cell potential for the Galvanic cell which is represented by the electrochemical reaction: \[ 2 \text{Cr(s)} + 3 \text{Fe}^{2+} (0.02M) \rightarrow 2 \text{Cr}^{3+} (0.2M) + 3 \text{Fe} \] \[ E^{0}_{\text{Fe}^{2+}/\text{Fe}} = -0.42 \, \text{V}, \quad E^{0}_{\text{Cr}^{3+}/\text{Cr}} = -0.72 \, \text{V} \]

• A. 0.3197 V
• B. 0.3364 V
• C. 0.2636 V
• D. 0.2803 V

Hint:

To calculate the cell potential under non-standard conditions, use the Nernst equation. Ensure you correctly assign oxidation and reduction reactions, and use the correct concentrations of the products and reactants.

Answer 1

The Correct Option is C

Step 1: Write the cell potential equation.
The cell potential \( E_{\text{cell}} \) is calculated using the Nernst equation for the half-reactions:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
where \( E_{\text{cathode}} \) is the reduction potential at the cathode and \( E_{\text{anode}} \) is the reduction potential at the anode.

Step 2: Assign the oxidation and reduction reactions.
- The chromium (Cr) is oxidized from \( \text{Cr(s)} \) to \( \text{Cr}^{3+} \), and this occurs at the anode.
- The iron (Fe) is reduced from \( \text{Fe}^{2+} \) to \( \text{Fe(s)} \), and this occurs at the cathode.
Thus, we assign:
- \( E_{\text{anode}} = E_{\text{Cr}^{3+}/\text{Cr}} = -0.72 \, \text{V} \) (oxidation of chromium),
- \( E_{\text{cathode}} = E_{\text{Fe}^{2+}/\text{Fe}} = -0.42 \, \text{V} \) (reduction of iron).

Step 3: Use the Nernst equation for non-standard conditions.
The Nernst equation is:
\[ E_{\text{cell}} = E^{0}_{\text{cell}} - \frac{0.0592}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
where:
- \( E^{0}_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \),
- \( n \) is the number of moles of electrons transferred in the reaction (which is 6 here, as 3 electrons are transferred per half-reaction),
- The concentrations of the products and reactants are \( [\text{Cr}^{3+}] = 0.2 \, \text{M} \) and \( [\text{Fe}^{2+}] = 0.02 \, \text{M} \).

Step 4: Calculate the standard cell potential \( E^{0}_{\text{cell}} \).
The standard cell potential is: \[ E^{0}_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = (-0.42 \, \text{V}) - (-0.72 \, \text{V}) = 0.30 \, \text{V} \]

Step 5: Substitute values into the Nernst equation.
Now substitute into the Nernst equation:
\[ E_{\text{cell}} = 0.30 \, \text{V} - \frac{0.0592}{6} \log \frac{0.2^2}{0.02^3} \]
\[ E_{\text{cell}} = 0.30 \, \text{V} - \frac{0.0592}{6} \log \frac{0.04}{8 \times 10^{-6}} = 0.30 \, \text{V} - \frac{0.0592}{6} \log 5000 \]
\[ E_{\text{cell}} = 0.30 \, \text{V} - 0.00987 \times 3.69897 \]
\[ E_{\text{cell}} = 0.30 \, \text{V} - 0.0365 \, \text{V} = 0.2636 \, \text{V} \]

Step 6: Conclusion.
Thus, the cell potential is \( 0.2636 \, \text{V} \), and the correct answer is option (C).