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: Recall the Nernst equation.
For any cell reaction the Nernst equation is \( E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{nF}\ln Q \), where \(n\) is the number of electrons transferred and \(Q\) is the reaction quotient (products over reactants, pure solids taken as 1).

Step 2: Find n.
Oxidation: \( Cr \rightarrow Cr^{3+} + 3e^- \), for 2 Cr this is 6 electrons.
Reduction: \( Cd^{2+} + 2e^- \rightarrow Cd \), for 3 Cd this is 6 electrons.
So \( n = 6 \).

Step 3: Write Q.
Solids Cr and Cd are omitted, so \( Q = \dfrac{[Cr^{3+}]^2}{[Cd^{2+}]^3} \).

Step 4: Substitute.
\( E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{6F}\ln\dfrac{[Cr^{3+}]^2}{[Cd^{2+}]^3} \), which is option (ii).

Why others are wrong: Options with \(+\) sign invert the correct thermodynamic form; option (iv) also uses a wrong \(n = 2\) and inverted Q.
\[\boxed{E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{6F}\ln\dfrac{[Cr^{3+}]^2}{[Cd^{2+}]^3}}\]