1. Identify the half-cells and standard reduction potentials

Given: E^_{Cr^{3+}/Cr} = -0.72\text{ V} E^_{Fe^{2+}/Fe} = -0.42\text{ V}

The cell is: $Cr\;|\;Cr^{3+}(0.1\,M)\;||\;Fe^{2+}(0.01\,M)\;|\;Fe$

So, left electrode is chromium and right electrode is iron.

2. Determine anode and cathode

The electrode with higher reduction potential undergoes reduction.

Since $-0.42\text{ V} > -0.72\text{ V}$ we have:

• Cathode: $Fe^{2+} + 2e^- \to Fe$
• Anode: $Cr \to Cr^{3+} + 3e^-$

3. Write the balanced overall cell reaction

Oxidation: $Cr \to Cr^{3+} + 3e^-$ Reduction: $Fe^{2+} + 2e^- \to Fe$

LCM of electrons $=6$, so: $2Cr \to 2Cr^{3+} + 6e^-$ $3Fe^{2+} + 6e^- \to 3Fe$

Overall reaction: $2Cr + 3Fe^{2+} \to 2Cr^{3+} + 3Fe$

Thus, $n=6$

4. Calculate standard cell potential

E^_{cell} = E^_{cathode} - E^_{anode} E^_{cell} = (-0.42) - (-0.72) = 0.30\text{ V}

5. Write the reaction quotient $Q$

From $2Cr + 3Fe^{2+} \to 2Cr^{3+} + 3Fe$

Solids are omitted, so: $Q = \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$

Given: $[Cr^{3+}] = 0.1, \quad [Fe^{2+}] = 0.01$

So, $Q = \frac{(0.1)^2}{(0.01)^3} = \frac{10^{-2}}{10^{-6}} = 10^4$

6. Apply Nernst equation

At $298\,K$: E_{cell} = E^_{cell} - \frac{0.0591}{n}\log Q

Substitute values: $E_{cell} = 0.30 - \frac{0.0591}{6}\log(10^4)$ $E_{cell} = 0.30 - \frac{0.0591}{6}\times 4$ $E_{cell} = 0.30 - 0.0394$ $E_{cell} \approx 0.2606\text{ V}$

Hence, $E_{cell} \approx 0.26\text{ V}$

7. Check options

• A: $0.26\text{ V}$ ✅
• B: $0.399\text{ V}$ ❌
• C: $-0.339\text{ V}$ ❌
• D: $-0.26\text{ V}$ ❌

Therefore, the correct option is A.