Question

For the cell reaction, \( 2Al(s) + 3Cu^{2+}_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3Cu(s) \), if \( \Delta G^\circ = -1158 \, \text{kJ} \), what is \( E_{\text{cell}} \)? 

• A. 3 V
• B. 2 V
• C. 2.5 V
• D. 1.5 V

Hint:

The standard cell potential can be calculated using the relation \( \Delta G^\circ = -nFE_{\text{cell}} \), with \( n \) being the number of electrons transferred in the reaction.

Answer 1

The Correct Option is B

Step 1: Understand the relationship between \( \Delta G^\circ \) and \( E_{\text{cell}} \).
The relationship between the Gibbs free energy change and the cell potential is given by:
\[ \Delta G^\circ = -nFE_{\text{cell}} \]
where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( n \) is the number of moles of electrons transferred,
- \( F \) is the Faraday constant (\( F = 96485 \, \text{C/mol} \)),
- \( E_{\text{cell}} \) is the standard cell potential.

Step 2: Calculate the number of moles of electrons transferred.
From the cell reaction:
\[ 2Al(s) + 3Cu^{2+}_{(aq)} \to 2Al^{3+}_{(aq)} + 3Cu(s) \]
We see that 6 electrons are transferred, since 2 moles of Al are oxidized (each Al loses 3 electrons) and 3 moles of Copper ion are reduced (each Copper ion gains 2 electrons). Therefore, \( n = 6 \) electrons.

Step 3: Rearrange the equation to solve for \( E_{\text{cell}} \).
Substitute the known values into the equation:
\[ \Delta G^\circ = -nFE_{\text{cell}} \quad \Rightarrow \quad E_{\text{cell}} = \frac{-\Delta G^\circ}{nF} \]
Substitute \( \Delta G^\circ = -1158 \, \text{kJ} = -1158 \times 10^3 \, \text{J} \), \( n = 6 \), and \( F = 96485 \, \text{C/mol} \): \[ E_{\text{cell}} = \frac{1158 \times 10^3}{6 \times 96485} = 2 \, \text{V} \]

Step 4: Final conclusion.
Thus, the standard cell potential is:
\[ \boxed{2 \, \text{V}} \]