The e.m.f of the following cell
`Ni(s)//NiSO_(4)(1.0M) ||H^(+)(1.0M)|H_(2)(1atm), Pt` at `25^(@)C` is `0.236V`. The electrical energy which can be product is

Options

A. `22.73 kJ`
B. `45.55 kJ`
C. `19.30 kJ`
D. `23.60kJ`

Explanation

To solve the problem, we need to calculate the electrical energy produced by the given electrochemical cell using the provided e.m.f (electromotive force) value. Here’s a step-by-step solution:

### Step 1: Understand the cell representation
The cell is represented as:
`Ni(s) | NiSO4(1.0M) || H+(1.0M) | H2(1 atm), Pt`

This indicates that nickel (Ni) is oxidized, and hydrogen ions (H+) are reduced to hydrogen gas (H2).

### Step 2: Write the cell reaction
From the cell representation, the overall cell reaction can be written as:
\[ \text{Ni(s)} + 2\text{H}^+ \rightarrow \text{NiSO}_4 + \text{H}_2(g) \]

### Step 3: Identify the number of electrons transferred (n)
In this reaction, each Ni atom donates 2 electrons when it is oxidized to Ni²⁺. Therefore, the number of electrons (n) participating in the reaction is:
\[ n = 2 \]

### Step 4: Use the Gibbs free energy equation
The relationship between Gibbs free energy (ΔG°) and the cell potential (E°) is given by:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \( F \) = Faraday's constant = 96500 C/mol
- \( E^\circ \) = e.m.f of the cell = 0.236 V

### Step 5: Substitute the values into the equation
Now we can substitute the values into the Gibbs free energy equation:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 0.236 \, \text{V} \]

### Step 6: Calculate ΔG°
Calculating the above expression:
\[ \Delta G^\circ = -2 \times 96500 \times 0.236 \]
\[ \Delta G^\circ = -45656 \, \text{J} \]
\[ \Delta G^\circ = -45.66 \, \text{kJ} \]

### Step 7: Interpret the result
The negative sign indicates that the reaction is spontaneous and the electrical energy produced by the cell is approximately:
\[ \Delta G^\circ \approx -45.66 \, \text{kJ} \]

### Final Answer
The electrical energy that can be produced is approximately **-45.66 kJ**.

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