Question

Given E⁰_Fe+3/Fe+2=+0.76V and E⁰½2l^-=+0.55V. The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is [\(\frac{2.303RT}{F}\)=0.06]

• A. 5x10¹²
• B. 1x10⁷
• C. 1x10⁹
• D. 3x10⁸

Answer 1

The Correct Option is B

The equilibrium constant \( K \) for a redox reaction can be related to the standard electrode potential difference \( \Delta E^\circ \) using the Nernst equation: \[ \Delta E^\circ = \frac{0.0592}{n} \log K \] where: - \( \Delta E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \) - \( n \) is the number of electrons involved in the reaction. For the given reaction, the electrode potentials are: \[ E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = + 0.76 \, \text{V} \quad \text{and} \quad E^\circ_{\text{I}_2/\text{I}^-} = + 0.55 \, \text{V} \] Thus, the potential difference \( \Delta E^\circ \) is: \[ \Delta E^\circ = 0.76 - 0.55 = 0.21 \, \text{V} \] The number of electrons \( n \) in the reaction is 2 (since both \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) and \( \text{I}_2 \) to \( \text{I}^- \) involve a 2-electron transfer). Substitute the values into the Nernst equation: \[ 0.21 = \frac{0.0592}{2} \log K \] Solving for \( K \): \[ \log K = \frac{0.21 \times 2}{0.0592} = 7.09 \] Thus, the equilibrium constant \( K \) is: \[ K = 10^{7.09} \approx 1 \times 10^7 \] Thus, the equilibrium constant is \( 1 \times 10^7 \), which corresponds to option (B).

The correct option is (B) : \(1\times10^7\)

Answer 2

The standard electrode potential for the given galvanic cell is calculated using the following formula for cell potential: \[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For this reaction, \( E_{\text{cell}} = 0.76 \, \text{V} - 0.55 \, \text{V} = 0.21 \, \text{V} \). Next, using the Nernst equation, the equilibrium constant \( K \) can be calculated as: \[ E_{\text{cell}} = \frac{0.0592}{n} \log K \] where \( n = 2 \), the number of electrons involved in the reaction. Solving for \( K \), we get: \[ K = 10^{\frac{n \cdot E_{\text{cell}}}{0.0592}} = 10^{\frac{2 \cdot 0.21}{0.0592}} = 1 \times 10^7 \]