Question

Calculate emf of the following cell at 298 K: \[ \text{Cr(s)} \, | \, \text{Cr}^{3+} (aq) \, (0.1\, M) \; || \; \text{Fe}^{2+} (aq) \, (0.01\, M) \, | \, \text{Fe(s)} \] Given: \[ E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, V, \quad E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, V, \quad \log 10 = 1 \]

Hint:

Steps for emf problems: 1. Identify anode and cathode 2. Balance electrons → find \( n \) 3. Calculate \( Q \) 4. Apply Nernst equation

Answer 1

Concept: The emf of a galvanic cell under non-standard conditions is calculated using the
Nernst equation: \[ E = E^\circ - \frac{0.0591}{n} \log Q \quad \text{(at 298 K)} \] Where:

• \( E^\circ \) = standard cell potential
• \( n \) = number of electrons transferred
• \( Q \) = reaction quotient

Step 1: Identify anode and cathode. Higher reduction potential → cathode. \[ E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 V>-0.74 V \] So:

• Cathode: Fe\(^{2+}\)/Fe
• Anode: Cr/Cr\(^{3+}\)

Step 2: Write half reactions. Anode (oxidation): \[ \text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \] Cathode (reduction): \[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \]
Step 3: Balance electrons. LCM of 3 and 2 = 6 electrons. Multiply: \[ 2(\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^-) \] \[ 3(\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}) \] Overall reaction: \[ 2\text{Cr} + 3\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 3\text{Fe} \] Thus, \( n = 6 \)
Step 4: Calculate standard emf. \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ = (-0.44) - (-0.74) = +0.30 \, V \]
Step 5: Calculate reaction quotient \( Q \). \[ Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} \] Substitute concentrations: \[ Q = \frac{(0.1)^2}{(0.01)^3} \] Convert to powers of 10: \[ 0.1 = 10^{-1}, \quad 0.01 = 10^{-2} \] \[ Q = \frac{(10^{-1})^2}{(10^{-2})^3} = \frac{10^{-2}}{10^{-6}} = 10^4 \]
Step 6: Apply Nernst equation. \[ E = 0.30 - \frac{0.0591}{6} \log(10^4) \] Since \( \log 10^4 = 4 \): \[ E = 0.30 - \frac{0.0591 \times 4}{6} \] \[ E = 0.30 - \frac{0.2364}{6} \] \[ E = 0.30 - 0.0394 \] \[ E \approx 0.26 \, V \]
Final Answer: \[ \therefore \text{Cell emf} \approx 0.26 \, V \]