Question

Consider the cell $\text{Cu} \mid \text{Cu}^{+2} \parallel \text{Ag}^{+} \mid \text{Ag}$. If the concentration of $\text{Cu}^{+2}$ and $\text{Ag}^{+}$ ions becomes ten times, then the emf of the cell will :-

• A. increases by 0.0591 V
• B. decreases by 0.0591 V
• C. increases by 0.0295 V
• D. decreases by 0.0295 V

Answer 1

Answer: (C)

increases by 0.0295 V

Answer 2

The given cell is $\text{Cu} \mid \text{Cu}^{2+} \parallel \text{Ag}^{+} \mid \text{Ag}$.

1. Half-cell reactions and the overall cell reaction:

   • Anode (oxidation): $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^{-}$
   • Cathode (reduction): $\text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s)$

   Balanced cathode reaction: $2\text{Ag}^{+}(aq) + 2e^{-} \rightarrow 2\text{Ag}(s)$

   Overall balanced cell reaction: $\text{Cu}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$

2. Number of electrons transferred (n): From the balanced reaction, $n = 2$.

3. Reaction quotient (Q): For the reaction $\text{Cu}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$, the reaction quotient $Q$ is: $Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^2}$

4. Nernst Equation: The Nernst equation at 298 K is: $E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0591}{n} \log Q$

   Initial concentrations: $[\text{Cu}^{2+}]_1$ and $[\text{Ag}^{+}]_1$. Initial EMF of the cell ($E_1$): $E_1 = E_{\text{cell}}^0 - \frac{0.0591}{2} \log \frac{[\text{Cu}^{2+}]_1}{[\text{Ag}^{+}]_1^2}$

5. New reaction quotient ($Q_2$) when concentrations become ten times: New concentrations: $[\text{Cu}^{2+}]_2 = 10[\text{Cu}^{2+}]_1$ and $[\text{Ag}^{+}]_2 = 10[\text{Ag}^{+}]_1$. The new reaction quotient ($Q_2$): $Q_2 = \frac{[\text{Cu}^{2+}]_2}{[\text{Ag}^{+}]_2^2} = \frac{10[\text{Cu}^{2+}]_1}{(10[\text{Ag}^{+}]_1)^2} = \frac{10[\text{Cu}^{2+}]_1}{100[\text{Ag}^{+}]_1^2} = \frac{1}{10} \frac{[\text{Cu}^{2+}]_1}{[\text{Ag}^{+}]_1^2}$ So, $Q_2 = \frac{1}{10} Q_1$.

6. New EMF ($E_2$): Substitute $Q_2$ into the Nernst equation: $E_2 = E_{\text{cell}}^0 - \frac{0.0591}{2} \log Q_2$ $E_2 = E_{\text{cell}}^0 - \frac{0.0591}{2} \log \left(\frac{1}{10} Q_1\right)$ $E_2 = E_{\text{cell}}^0 - \frac{0.0591}{2} \left(\log \frac{1}{10} + \log Q_1\right)$ Since $\log \frac{1}{10} = -1$: $E_2 = E_{\text{cell}}^0 - \frac{0.0591}{2} \left(-1 + \log Q_1\right)$ $E_2 = E_{\text{cell}}^0 - \frac{0.0591}{2} \log Q_1 + \frac{0.0591}{2}$ $E_2 = \left(E_{\text{cell}}^0 - \frac{0.0591}{2} \log Q_1\right) + \frac{0.0591}{2}$ We know that the term in the parenthesis is the initial EMF, $E_1$: $E_2 = E_1 + \frac{0.0591}{2}$

7. Change in EMF: The change in EMF is $E_2 - E_1$: $\Delta E = E_2 - E_1 = \frac{0.0591}{2}$ $\Delta E = 0.02955 \text{ V}$

   Thus, the emf of the cell increases by 0.0295 V.