Question

Which of the following expressions is used to find the cell potential of $Cd_{(s)} | Cd_{(aq)}^{2+} \, || \, Cu_{(aq)}^{2+} | Cu_{(s)}$ cell at $25^\circ \text{C}$?

• A. $E_{cell}=E_{cell}^{o}-0.0296~\log\frac{[Cd^{++}]}{[Cu^{++}]}$
• B. $E_{cell}=E_{cell}^{o}+0.0296~\log\frac{[Cd^{++}]}{[Cu^{++}]}$
• C. $E_{cell}=E_{cell}^{o}-0.0592~\log\frac{[Cu^{++}]}{[Cd^{++}]}$
• D. $E_{cell}=E_{cell}^{o}+0.0592~\log\frac{[Cu^{++}]}{[Cd^{++}]}$

Hint:

Chemistry Tip: In the Nernst equation $E = E^\circ - \frac{0.0592}{n} \log Q$, always remember that $Q$ is $[\text{Anode}]/[\text{Cathode}]$ for standard galvanic cells with metal electrodes.

Answer 1

The Correct Option is A

Concept: Chemistry (Electrochemistry) - Nernst Equation.

Step 1: Identify the half-cell reactions. From the given cell notation $Cd_{(s)}|Cd_{(aq)}^{++}||Cu_{(aq)}^{++}|Cu_{(s)}$, we can write the half-reactions. At the anode (oxidation): $Cd_{(s)} \rightarrow Cd_{(aq)}^{2+} + 2e^-$. At the cathode (reduction): $Cu_{(aq)}^{2+} + 2e^- \rightarrow Cu_{(s)}$.

Step 2: Determine the overall reaction and number of electrons transferred. Combining the half-reactions gives the overall cell reaction: $Cd_{(s)} + Cu_{(aq)}^{2+} \rightarrow Cd_{(aq)}^{2+} + Cu_{(s)}$. The number of moles of electrons transferred in this balanced reaction is $n = 2$.

Step 3: State the standard Nernst equation. The Nernst equation for a cell potential at $25^{\circ}C$ (298 K) is given by: $E_{cell} = E_{cell}^{o} - \frac{0.0592}{n} \log_{10} Q$, where $Q$ is the reaction quotient.

Step 4: Formulate the reaction quotient $Q$. The reaction quotient $Q$ is the ratio of the concentrations of the aqueous products to the aqueous reactants. Solid concentrations are taken as unity. $Q = \frac{[Cd^{2+}]}{[Cu^{2+}]}$.

Step 5: Substitute values into the Nernst equation. Substitute $n = 2$ and $Q$ into the equation: $E_{cell} = E_{cell}^{o} - \frac{0.0592}{2} \log_{10} \frac{[Cd^{2+}]}{[Cu^{2+}]}$. Simplifying the fraction $\frac{0.0592}{2}$ gives $0.0296$. The final expression is $E_{cell} = E_{cell}^{o} - 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$. This perfectly matches option A. $$ \therefore \text{The correct expression is } E_{cell}=E_{cell}^{o}-0.0296~\log\frac{[Cd^{++}]}{[Cu^{++}]}. $$