Question

The standard emf for cell, Cd$_{(s)}$ | Cd$^{+2}$(1M) || Cu$^{+2}$(1M) | Cu$_{(s)}$ is 0.74 V.
If concentration of Cd$^{+2}_{(aq)}$ and Cu$^{+2}_{(aq)}$ decreases by 10 times at 298 K. Calculate emf of cell.

• A. +0.074 V
• B. +0.850 V
• C. +0.680 V
• D. +0.740 V

Hint:

If the concentration of both anode and cathode ions changes by the same factor in a symmetrical cell, the cell EMF remains equal to the standard EMF.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question provides the standard EMF (E°\(_\text{cell}\)) of a galvanic cell and asks for the new EMF (E\(_\text{cell}\)) when the concentrations of both ions are changed by the same factor. We need to use the Nernst equation to solve this.
Step 2: Key Formula or Approach:
The Nernst equation relates the cell potential (E\(_\text{cell}\)) to the standard cell potential (E°\(_\text{cell}\)) and the reaction quotient (Q):
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log_{10} Q \quad (\text{at 298 K}) \] First, we need to write the overall cell reaction to determine the number of electrons transferred (n) and the expression for Q.
Step 3: Detailed Explanation:
The cell notation is Cd\(_{(s)}\) | Cd\(^{+2}\) || Cu\(^{+2}\) | Cu\(_{(s)}\).
Oxidation at Anode: Cd\(_{(s)}\) \( \rightarrow \) Cd\(^{+2}_{(aq)}\) + 2e\(^-\)
Reduction at Cathode: Cu\(^{+2}_{(aq)}\) + 2e\(^-\) \( \rightarrow \) Cu\(_{(s)}\)
Overall Cell Reaction: Cd\(_{(s)}\) + Cu\(^{+2}_{(aq)}\) \( \rightarrow \) Cd\(^{+2}_{(aq)}\) + Cu\(_{(s)}\)
From the balanced reaction, the number of electrons transferred, n = 2.
The reaction quotient, Q, is given by:
\[ Q = \frac{[\text{Products}]}{[\text{Reactants}]} = \frac{[\text{Cd}^{+2}]}{[\text{Cu}^{+2}]} \] (Note: The concentrations of pure solids Cd\(_{(s)}\) and Cu\(_{(s)}\) are taken as 1).
Initial Condition (Standard state):
[Cd\(^{+2}\)] = 1 M, [Cu\(^{+2}\)] = 1 M.
\(Q = \frac{1}{1} = 1\).
\(E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log(1) = E^\circ_{\text{cell}} - 0 = E^\circ_{\text{cell}}\).
Given E°\(_\text{cell}\) = 0.74 V.
New Condition:
The concentrations of both Cd\(^{+2}\) and Cu\(^{+2}\) decrease by 10 times.
New [Cd\(^{+2}\)] = 1 M / 10 = 0.1 M.
New [Cu\(^{+2}\)] = 1 M / 10 = 0.1 M.
Now, calculate the new reaction quotient, Q':
\[ Q' = \frac{[\text{New Cd}^{+2}]}{[\text{New Cu}^{+2}]} = \frac{0.1 \text{ M}}{0.1 \text{ M}} = 1 \] Calculate the new cell EMF using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log_{10} Q' \] \[ E_{\text{cell}} = 0.74 \text{ V} - \frac{0.0591}{2} \log_{10}(1) \] Since log\(_\text{10}\)(1) = 0:
\[ E_{\text{cell}} = 0.74 \text{ V} - 0 = 0.74 \text{ V} \] Step 4: Final Answer:
The emf of the cell remains unchanged at +0.740 V because the ratio of the ion concentrations (the reaction quotient Q) did not change.