The standard emf for the cell cell reaction ` Zn + Cu^(2+) rarr Zn^(2+) + Cu ` is 1.10 volt at ` 25^@ C`. The emf for the cell reaction when ` 0.1 M Cu^(2+)` and ` 0.1 M ZN^(2+)` solutions are used at `25^@ =C` is .

Options

A. `1.10` volt
B. `0.110` volt
C. `-1.10` volt
D. `-0.110` volt

Explanation

To find the EMF of the cell reaction when 0.1 M Cu²⁺ and 0.1 M Zn²⁺ solutions are used at 25°C, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write down the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
where:
- \( E_{\text{cell}} \) is the cell potential under non-standard conditions,
- \( E^\circ_{\text{cell}} \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred in the reaction,
- \([\text{Products}]\) and \([\text{Reactants}]\) are the concentrations of the products and reactants.

### Step 2: Identify the standard cell potential and number of electrons transferred
From the question, we know:
- \( E^\circ_{\text{cell}} = 1.10 \, \text{V} \)
- The reaction is:
\[ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \]
- Here, Zn is oxidized (loses 2 electrons) and Cu²⁺ is reduced (gains 2 electrons), so \( n = 2 \).

### Step 3: Determine the concentrations of products and reactants
In this case, the products are \([\text{Zn}^{2+}]\) and \([\text{Cu}]\), and the reactants are \([\text{Cu}^{2+}]\) and \([\text{Zn}]\). Given that:
- \([\text{Cu}^{2+}] = 0.1 \, \text{M}\)
- \([\text{Zn}^{2+}] = 0.1 \, \text{M}\)
- The concentrations of pure solids (Zn and Cu) are considered to be 1.

Thus, we can write:
\[ [\text{Products}] = [\text{Zn}^{2+}] \cdot [\text{Cu}] = 0.1 \cdot 1 = 0.1 \]
\[ [\text{Reactants}] = [\text{Cu}^{2+}] \cdot [\text{Zn}] = 0.1 \cdot 1 = 0.1 \]

### Step 4: Substitute values into the Nernst equation
Now substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \log \left( \frac{0.1}{0.1} \right) \]

### Step 5: Simplify the logarithm
Since \(\log(1) = 0\):
\[ E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \cdot 0 \]
\[ E_{\text{cell}} = 1.10 - 0 \]
\[ E_{\text{cell}} = 1.10 \, \text{V} \]

### Conclusion
The EMF for the cell reaction when 0.1 M Cu²⁺ and 0.1 M Zn²⁺ solutions are used at 25°C is **1.10 V**.

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