Calculate emf of the following cell reaction at 2968 K :
`Ni(s)//Ni^(2+)(0.01 M) || Cu^(2+)(0.1 M)//Cu(s)`
[Given `E_(Ni^(2+)//Ni)^(@)=-0.25" V ",E_(Cu^(2+)//Cu)^(@)=+0.34 " V "`]
Write the overall cell reaction.

Explanation

To calculate the EMF of the given cell reaction at 298 K, we will follow these steps:

### Step 1: Identify the half-reactions
The cell consists of two half-reactions:
1. Nickel half-reaction: \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \) with \( E^\circ = -0.25 \, \text{V} \)
2. Copper half-reaction: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^\circ = +0.34 \, \text{V} \)

### Step 2: Write the overall cell reaction
The overall cell reaction can be obtained by combining the two half-reactions. Since both half-reactions involve 2 electrons, they can be directly added:
\[ \text{Ni}^{2+} (aq) + \text{Cu} (s) \rightarrow \text{Ni} (s) + \text{Cu}^{2+} (aq) \]

### Step 3: Calculate the standard cell potential (\( E^\circ_{cell} \))
The standard cell potential can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Here, the cathode is the copper half-reaction (reduction) and the anode is the nickel half-reaction (oxidation):
\[ E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Ni^{2+}/Ni} = 0.34 \, \text{V} - (-0.25 \, \text{V}) = 0.34 + 0.25 = 0.59 \, \text{V} \]

### Step 4: Use the Nernst equation to calculate the EMF at non-standard conditions
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( n = 2 \) (number of electrons transferred)
- \( F = 96485 \, \text{C/mol} \)
- \( Q = \frac{[\text{Ni}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.01}{0.1} = 0.1 \)

### Step 5: Calculate \( E_{cell} \)
Substituting the values into the Nernst equation:
\[ E_{cell} = 0.59 - \frac{(8.314)(298)}{(2)(96485)} \ln(0.1) \]
Calculating the term:
\[ \frac{(8.314)(298)}{(2)(96485)} \approx 0.00414 \, \text{V} \]
Now, calculate \( \ln(0.1) \):
\[ \ln(0.1) \approx -2.302 \]
Now substituting back:
\[ E_{cell} = 0.59 - (0.00414)(-2.302) \approx 0.59 + 0.00952 \approx 0.59952 \, \text{V} \approx 0.60 \, \text{V} \]

### Final Answer
The EMF of the cell reaction at 298 K is approximately **0.60 V**.