Calculate the `EMF` of the cell in whiCHM the following reaction takes place `:`
`Ni(s)+2Ag^(o+)(0.002M) rarr Ni^(2+)(0.160M)+2Ag(s)`

Explanation

To calculate the EMF of the cell for the reaction:

\[ \text{Ni(s)} + 2\text{Ag}^+(0.002M) \rightarrow \text{Ni}^{2+}(0.160M) + 2\text{Ag(s)} \]

we will follow these steps:

### Step 1: Identify the standard EMF of the cell
The standard EMF (\(E^\circ\)) for the cell reaction is given as 1.05 volts.

### Step 2: Determine the number of electrons transferred (N)
In the given reaction, 2 electrons are transferred (as 2 moles of Ag\(^+\) are reduced to 2 moles of Ag). Therefore, \(N = 2\).

### Step 3: Write the Nernst equation
The Nernst equation for calculating the EMF of the cell is given by:

\[ E = E^\circ - \frac{0.0591}{N} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

### Step 4: Identify the concentrations of products and reactants
From the reaction:
- Products: \([\text{Ni}^{2+}] = 0.160M\) and \([\text{Ag}] = 1\) (since solid Ag is not included in the expression)
- Reactants: \([\text{Ag}^+] = 0.002M\) and \([\text{Ni}] = 1\) (since solid Ni is not included in the expression)

### Step 5: Substitute values into the Nernst equation
Substituting the values into the Nernst equation:

\[ E = 1.05 - \frac{0.0591}{2} \log \left( \frac{[0.160]}{[0.002]^2} \right) \]

### Step 6: Calculate the ratio of concentrations
Calculate the ratio of concentrations:

\[ \frac{[0.160]}{[0.002]^2} = \frac{0.160}{0.000004} = 40000 \]

### Step 7: Calculate the logarithm
Now calculate the logarithm:

\[ \log(40000) = \log(4 \times 10^4) = \log(4) + \log(10^4) = 0.6021 + 4 = 4.6021 \]

### Step 8: Substitute the logarithm back into the equation
Substituting back into the Nernst equation gives:

\[ E = 1.05 - \frac{0.0591}{2} \times 4.6021 \]

### Step 9: Perform the calculation
Calculate the second term:

\[ \frac{0.0591}{2} \times 4.6021 = 0.02755 \times 4.6021 \approx 0.1274 \]

Now substitute this back into the equation:

\[ E = 1.05 - 0.1274 = 0.9226 \approx 0.91 \text{ volts} \]

### Final Answer
The EMF of the cell is approximately **0.91 volts**.

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