For the redox reaction `Zn(s) + Cu^(2+) (0.1M) rarr Zn^(2+) (1M) + Cu(s)` that takes place in a cell, `E^(o)""_(cell)` is `1.10` volt. `E_(cell)` for the cell will be:

Options

A. `2.14` volt
B. `1.80` volt
C. `1.07` volt
D. `1.82` volt

Answer

To find the cell potential \( E_{cell} \) for the given redox reaction \( \text{Zn(s)} + \text{Cu}^{2+} (0.1M) \rightarrow \text{Zn}^{2+} (1M) + \text{Cu(s)} \), we will use the Nernst equation:

\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]

### Step 1: Identify the standard cell potential
The standard cell potential \( E^{\circ}_{cell} \) is given as \( 1.10 \, \text{V} \).

### Step 2: Determine the number of electrons transferred (n)
In the reaction:
- Zinc (\( \text{Zn} \)) is oxidized from \( \text{Zn}^0 \) to \( \text{Zn}^{2+} \), losing 2 electrons.
- Copper (\( \text{Cu}^{2+} \)) is reduced from \( \text{Cu}^{2+} \) to \( \text{Cu}^0 \), gaining 2 electrons.

Thus, the number of electrons transferred \( n = 2 \).

### Step 3: Write the concentrations of products and reactants
- The concentration of the product \( [\text{Zn}^{2+}] = 1 \, \text{M} \)
- The concentration of the reactant \( [\text{Cu}^{2+}] = 0.1 \, \text{M} \)

### Step 4: Substitute values into the Nernst equation
Now we can substitute the values into the Nernst equation:

\[ E_{cell} = 1.10 - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right) \]

### Step 5: Simplify the logarithm
The logarithm can be simplified:

\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

### Step 6: Substitute the logarithm value back into the equation
Now substitute this back into the equation:

\[ E_{cell} = 1.10 - \frac{0.0591}{2} \cdot 1 \]

### Step 7: Calculate the value
Calculating the term:

\[ \frac{0.0591}{2} = 0.02955 \]

Now, substituting this value:

\[ E_{cell} = 1.10 - 0.02955 = 1.07045 \, \text{V} \]

### Step 8: Final answer
Rounding to two decimal places, we get:

\[ E_{cell} \approx 1.07 \, \text{V} \]

Thus, the cell potential \( E_{cell} \) for the reaction is approximately \( 1.07 \, \text{V} \).

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