The standard oxidation potential of `Zn` referred to SHE is `0.76V` and that of `Cu` is `-0.34V` at `25^(@)C`. When excess of `Zn` is added to `CuSO_(4),Zn` diplaces `Cu^(2+)` till equilibrium is reached. What is the ratio of `Zn^(2+)` to `Cu^(2+)` ions at equilibrium?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.
The half-reactions for zinc and copper are:
1. **Zinc oxidation**: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) (Standard oxidation potential of Zn = 0.76 V)
2. **Copper reduction**: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) (Standard reduction potential of Cu = -0.34 V)

### Step 2: Calculate the standard cell potential (\(E^\circ_{\text{cell}}\)).
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is where reduction occurs (Cu) and the anode is where oxidation occurs (Zn). Thus:
\[ E^\circ_{\text{cell}} = (-0.34 \, \text{V}) - (0.76 \, \text{V}) = -0.34 + 0.76 = 1.1 \, \text{V} \]

### Step 3: Write the Nernst equation.
The Nernst equation relates the cell potential to the concentrations of the reactants and products:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
Where \( n \) is the number of moles of electrons transferred (which is 2 in this case).

### Step 4: Set up the equation at equilibrium.
At equilibrium, \( E_{\text{cell}} = 0 \):
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
Rearranging gives:
\[ \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = E^\circ_{\text{cell}} \]

### Step 5: Substitute the value of \( E^\circ_{\text{cell}} \).
Substituting \( E^\circ_{\text{cell}} = 1.1 \, \text{V} \):
\[ \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = 1.1 \]

### Step 6: Solve for the ratio of concentrations.
Multiplying both sides by \( \frac{2}{0.0591} \):
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) = \frac{1.1 \times 2}{0.0591} \]
Calculating the right side:
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \approx 37.288 \]

### Step 7: Take the antilog to find the ratio.
Taking the antilogarithm:
\[ \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 10^{37.288} \approx 1.941 \times 10^{37} \]

### Final Answer:
The ratio of \( [\text{Zn}^{2+}] \) to \( [\text{Cu}^{2+}] \) at equilibrium is approximately \( 1.941 \times 10^{37} \).

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