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: Understanding the Question:
We need to calculate the standard cell potential (\( \text{E}^\circ_{\text{cell}} \)) using the standard Gibbs free energy change (\( \Delta\text{G}^\circ \)).
Step 2: Key Formula or Approach:
Use the relationship:
\[ \Delta\text{G}^\circ = -n\text{F}\text{E}^\circ_{\text{cell}} \] where \( n \) is the number of moles of electrons transferred and \( \text{F} \) is Faraday's constant (\( \approx 96500 \text{ C/mol} \)).
Step 3: Detailed Explanation:
1. Determine \( n \) from the balanced equation:
Oxidation: \( \text{2Al} \longrightarrow 2\text{Al}^{3+} + 6\text{e}^- \)
Reduction: \( 3\text{Cu}^{2+} + 6\text{e}^- \longrightarrow 3\text{Cu} \)
Total electrons transferred, \( n = 6 \).
2. Given \( \Delta\text{G}^\circ = -1158 \text{ kJ} = -1158000 \text{ J} \).
3. Calculate \( \text{E}^\circ_{\text{cell}} \):
\[ -1158000 = - (6) \times (96500) \times \text{E}^\circ_{\text{cell}} \] \[ \text{E}^\circ_{\text{cell}} = \frac{1158000}{579000} \] \[ \text{E}^\circ_{\text{cell}} = 2 \text{ V} \] Step 4: Final Answer:
The standard cell potential is 2 V.