Consider the following reaction at equilibrium `2Fe^(2+)(aq)+Cu^(2+)to2Fe^(3+)(aq)+Cu`
When the reaction comes to equilibrium, what is the cell voltage ?

Options

A. 0.43 V
B. 1.11 V
C. 0.78 V
D. 0 V

Answer

To find the cell voltage (E_cell) for the given reaction at equilibrium, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ 2Fe^{2+}(aq) + Cu^{2+} \rightleftharpoons 2Fe^{3+}(aq) + Cu \]
This indicates that iron(II) ions are being oxidized to iron(III) ions while copper(II) ions are being reduced to copper metal.

### Step 2: Recognize the Concept of Equilibrium
At equilibrium, the forward and reverse reactions occur at the same rate, meaning that the concentrations of the reactants and products remain constant.

### Step 3: Relate Gibbs Free Energy to Cell Voltage
The relationship between Gibbs free energy (ΔG) and cell voltage (E_cell) is given by the equation:
\[ \Delta G = -nFE_{cell} \]
Where:
- ΔG = Change in Gibbs free energy
- n = Number of moles of electrons transferred in the reaction
- F = Faraday's constant (approximately 96485 C/mol)
- E_cell = Cell voltage

### Step 4: Analyze the Condition at Equilibrium
At equilibrium, the change in Gibbs free energy (ΔG) is zero:
\[ \Delta G = 0 \]
Substituting this into the equation gives:
\[ 0 = -nFE_{cell} \]
This implies:
\[ E_{cell} = 0 \]

### Step 5: Conclusion
Since the cell voltage at equilibrium is zero, we conclude:
\[ E_{cell} = 0 \, V \]

### Final Answer
The cell voltage when the reaction comes to equilibrium is:
\[ \text{E}_{cell} = 0 \, V \]

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