The emf of the cell in which the following reactions,
`Zn(s)+Ni^(2+) (0.1M) rarr Zn^(2+) (1.0 M) + Ni(s)` occurs, is found to 0.5105 V at 298 K. The standard emf of the cell is :

Options

A. `0.4810 V`
B. `0.5696 V`
C. `-0.5105 V`
D. `0.5400 V`

Explanation

To find the standard EMF of the cell given the reaction and the EMF at specific conditions, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[Ox]}{[Red]} \right) \]
Where:
- \(E\) = EMF of the cell
- \(E^\circ\) = Standard EMF of the cell
- \(n\) = number of moles of electrons transferred
- \([Ox]\) = concentration of oxidized species
- \([Red]\) = concentration of reduced species

### Step 2: Identify the Components
From the given reaction:
\[ \text{Zn(s)} + \text{Ni}^{2+} (0.1M) \rightarrow \text{Zn}^{2+} (1.0M) + \text{Ni(s)} \]
- The oxidized species is \(\text{Zn}^{2+}\) and the reduced species is \(\text{Ni}^{2+}\).
- The concentration of \(\text{Zn}^{2+}\) is 1.0 M and the concentration of \(\text{Ni}^{2+}\) is 0.1 M.

### Step 3: Determine the Number of Electrons Transferred
In this reaction, zinc is oxidized to zinc ions, and nickel ions are reduced to nickel metal. The half-reactions are:
- \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) (oxidation)
- \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \) (reduction)

Thus, \(n = 2\).

### Step 4: Substitute Values into the Nernst Equation
We know:
- \(E = 0.5105 \, V\)
- \([Ox] = [Zn^{2+}] = 1.0 \, M\)
- \([Red] = [Ni^{2+}] = 0.1 \, M\)

Substituting these values into the Nernst equation:
\[ 0.5105 = E^\circ - \frac{0.0591}{2} \log \left( \frac{1.0}{0.1} \right) \]

### Step 5: Calculate the Logarithm
Calculate the logarithm:
\[ \log \left( \frac{1.0}{0.1} \right) = \log(10) = 1 \]

### Step 6: Substitute and Solve for \(E^\circ\)
Now substitute back into the equation:
\[ 0.5105 = E^\circ - \frac{0.0591}{2} \times 1 \]
\[ 0.5105 = E^\circ - 0.02955 \]
\[ E^\circ = 0.5105 + 0.02955 \]
\[ E^\circ = 0.54005 \, V \]

### Step 7: Round to Appropriate Significant Figures
Rounding to four significant figures:
\[ E^\circ \approx 0.540 \, V \]

### Final Answer
The standard EMF of the cell is approximately **0.540 V**.

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