Calculate the maximum electrical work that can be obtained from the following cell under the standard conditions at `25^(@)C`.
`Zn//Zn^(2+) (aq)"||"Fe^(2+)(aq)//Fe`
Given that `E_(Zn^(2+)(aq)//Zn)^(Theta)= -0.76V`
`E^(Theta)Fe^(2+) (aq)//Fe= -0.44 V and F = 96500"C mol"^(-1)`

Explanation

To solve the problem of calculating the maximum electrical work that can be obtained from the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
The cell is represented as:
\[ \text{Zn} | \text{Zn}^{2+} (aq) || \text{Fe}^{2+} (aq) | \text{Fe} \]

From the problem, we have:
- For the zinc half-reaction:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad (E^\circ = -0.76 \, \text{V}) \]
- For the iron half-reaction:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ = -0.44 \, \text{V}) \]

### Step 2: Determine the anode and cathode.
- The anode is where oxidation occurs (Zn is oxidized).
- The cathode is where reduction occurs (Fe²⁺ is reduced).

### Step 3: Calculate the standard cell potential (\(E^\circ_{\text{cell}}\)).
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.44 \, \text{V}) - (-0.76 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.44 + 0.76 = 0.32 \, \text{V} \]

### Step 4: Calculate the number of moles of electrons transferred (\(n\)).
From the half-reactions, we see that 2 moles of electrons are transferred during the reaction:
\[ n = 2 \]

### Step 5: Use the formula for Gibbs free energy change (\(\Delta G\)).
The relationship between the maximum electrical work and Gibbs free energy is given by:
\[ \Delta G = -nFE^\circ_{\text{cell}} \]
Where:
- \(F = 96500 \, \text{C/mol}\) (Faraday's constant)

Substituting the values:
\[ \Delta G = -2 \times 96500 \, \text{C/mol} \times 0.32 \, \text{V} \]
\[ \Delta G = -2 \times 96500 \times 0.32 \]
\[ \Delta G = -61,760 \, \text{J} \]
Converting to kilojoules:
\[ \Delta G = -61.76 \, \text{kJ} \]

### Conclusion:
The maximum electrical work that can be obtained from the cell under standard conditions at \(25^\circ C\) is \(-61.76 \, \text{kJ}\).

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