The EMF of the cell `Ni|Ni^(2+)` (0.01M)
`Cl^(-)(0.01M)//Cl_(2)`, pt is__ V if the SRP of nickel and chlorine electrodes are `-0.25` and `+1.36 V` respectively

Options

A. `+1.61`
B. `-1.61`
C. `+1.79`
D. `-1.79`

Explanation

To find the EMF of the cell `Ni|Ni^(2+)` (0.01M) `Cl^(-)(0.01M)//Cl_(2)`, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials (SRP)
- The reduction half-reaction for nickel is:
\[ Ni^{2+} + 2e^- \rightarrow Ni \quad (E^\circ = -0.25 \, V) \]
- The reduction half-reaction for chlorine is:
\[ Cl_2 + 2e^- \rightarrow 2Cl^- \quad (E^\circ = +1.36 \, V) \]

### Step 2: Determine which half-reaction will be oxidation and which will be reduction
- Since chlorine has a higher reduction potential (+1.36 V) compared to nickel (-0.25 V), chlorine will undergo reduction and nickel will undergo oxidation.

### Step 3: Write the overall cell reaction
- The oxidation half-reaction for nickel is:
\[ Ni \rightarrow Ni^{2+} + 2e^- \quad (E^\circ = +0.25 \, V) \]
- The overall cell reaction combining both half-reactions is:
\[ Ni + Cl_2 \rightarrow Ni^{2+} + 2Cl^- \]

### Step 4: Calculate the standard cell potential (E°cell)
- The standard cell potential is given by:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
- Here, the cathode is chlorine and the anode is nickel:
\[ E^\circ_{cell} = 1.36 \, V - (-0.25 \, V) = 1.36 \, V + 0.25 \, V = 1.61 \, V \]

### Step 5: Use the Nernst equation to calculate the cell EMF
- The Nernst equation is:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
- Where:
- \( n = 2 \) (number of electrons transferred)
- \( Q = \frac{[Ni^{2+}]}{[Cl^-]^2} \) (reaction quotient)
- Given concentrations: \([Ni^{2+}] = 0.01 \, M\) and \([Cl^-] = 0.01 \, M\)
- Therefore, \( Q = \frac{0.01}{(0.01)^2} = \frac{0.01}{0.0001} = 100 \)

### Step 6: Substitute values into the Nernst equation
- Now substituting into the Nernst equation:
\[ E_{cell} = 1.61 \, V - \frac{0.0591}{2} \log(100) \]
- Since \( \log(100) = 2 \):
\[ E_{cell} = 1.61 \, V - \frac{0.0591}{2} \times 2 \]
\[ E_{cell} = 1.61 \, V - 0.0591 \]
\[ E_{cell} = 1.61 \, V - 0.02955 = 1.58045 \, V \]

### Step 7: Final calculation
- Rounding off, we get:
\[ E_{cell} \approx 1.58 \, V \]

### Conclusion
The EMF of the cell is approximately **1.58 V**.

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