Question

What will be the Nernst equation of the cell for the following reaction?
\( 2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s) \)

• A. \( E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{6F}\ln\dfrac{[Cd^{2+}]^3}{[Cr^{3+}]^2} \)
• B. \( E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{6F}\ln\dfrac{[Cr^{3+}]^2}{[Cd^{2+}]^3} \)
• C. \( E_{cell} = E^{\circ}_{cell} + \dfrac{RT}{6F}\ln\dfrac{[Cr^{3+}]^2}{[Cd^{2+}]^3} \)
• D. \( E_{cell} = E^{\circ}_{cell} + \dfrac{RT}{2F}\ln\dfrac{[Cd^{2+}]^3}{[Cr^{3+}]^2} \)

Hint:

Count electrons transferred (n = 6) and write Q as products over reactants, then apply \( E = E^{\circ} - \frac{RT}{nF}\ln Q \).

Answer 1

The Correct Option is B

Step 1: Split into half-reactions and balance electrons.
Each Cr gives 3 electrons and each Cd\(^{2+}\) takes 2 electrons. The least common multiple of 3 and 2 is 6, so 2 Cr and 3 Cd\(^{2+}\) exchange exactly 6 electrons, giving \(n = 6\).

Step 2: Build the reaction quotient.
\(Q\) is written as concentration of products raised to their coefficients divided by that of reactants; solids have activity 1. Hence \(Q = \dfrac{[Cr^{3+}]^2}{[Cd^{2+}]^3}\).

Step 3: Insert into the Nernst form.
Using \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{nF}\ln Q\) with \(n = 6\) gives \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{6F}\ln\dfrac{[Cr^{3+}]^2}{[Cd^{2+}]^3}\).

Step 4: Match the option.
This is exactly option (ii). The sign is negative because increasing product ion concentration lowers the cell potential.
\[\boxed{\text{Option (ii)}}\]