Question

Given that: \[ E^\circ_{\text{Fe}^{3+}/\text{Fe}} = -0.036\, \text{V} \quad \text{and} \quad E^\circ_{\text{M}^{x+}/\text{M}} = 0.15\, \text{V} \] A Galvanic cell is formed using above electrodes, whose \( E_{\text{cell}} = 0.2047\, \text{V} \) when reaction quotient of cell reaction is \(10^{-2}\). Find the value of \(x\). [Nearest integer]

• A. \(1 \)
• B. \(2 \)
• C. \(3 \)
• D. \(4 \)

Answer 1

The Correct Option is B

Concept: The EMF of a galvanic cell under non-standard conditions is given by the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q \] where: • \(E^\circ_{\text{cell}}\) is standard EMF, • \(n\) is number of electrons transferred, • \(Q\) is reaction quotient.
Step 1: Write the half-cell reactions.
Anode (oxidation): \[ \text{Fe}(s) \rightarrow \text{Fe}^{3+}(aq) + 3e^- \] Cathode (reduction): \[ \text{M}^{x+}(aq) + xe^- \rightarrow \text{M}(s) \]
Step 2: Balance the electrons.
LCM of electrons = \(3x\) Overall reaction: \[ 3\text{M}^{x+} + x\text{Fe}(s) \rightarrow 3\text{M}(s) + x\text{Fe}^{3+} \]
Step 3: Calculate standard EMF.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = 0.15 - (-0.036) = 0.186\, \text{V} \]
Step 4: Apply Nernst equation.
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{3x} \log(10^{-2}) \] \[ 0.2047 = 0.186 - \frac{0.059}{3x}(-2) \] \[ 0.2047 = 0.186 + \frac{0.118}{3x} \]
Step 5: Solve for \(x\).
\[ 0.2047 - 0.186 = \frac{0.118}{3x} \] \[ 0.0187 = \frac{0.118}{3x} \] \[ 3x = \frac{0.118}{0.0187} \approx 6.31 \] \[ x \approx 2.1 \] Nearest integer: \[ x = 2 \]