To find the theoretical ΔG for the copper-zinc voltaic cell, we can use the formula:

$\Delta G = -nFE_{\text{cell}}$

where:

• $n$ is the number of moles of electrons transferred,
• $F$ is Faraday's constant ($F = 96500 \, \text{J/V·mol}$),
• $E_{\text{cell}}$ is the cell potential.

Step 1: Determine the Cell Potential, $E_{\text{cell}}$

For a voltaic cell, the cell potential $E_{\text{cell}}$ is the difference in the reduction potentials of the two half-reactions.

The half-reactions are:

1. Oxidation at the anode (zinc): $\text{Zn(s)} \rightarrow \text{Zn}^{2+} + 2e^-$ The standard reduction potential for zinc ($\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn(s)}$) is approximately $-0.76 \, \text{V}$.

2. Reduction at the cathode (copper): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)}$ The standard reduction potential for copper ($\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)}$) is approximately $+0.34 \, \text{V}$.

The cell potential $E_{\text{cell}}$ is calculated as: $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 1.10 \, \text{V}$

Step 2: Calculate $\Delta G$

Since 2 moles of electrons are transferred ($n = 2$), we can now calculate $\Delta G$:

$\Delta G = -nFE_{\text{cell}} = -(2)(96500 \, \text{J/V·mol})(1.10 \, \text{V})$

$\Delta G = -212300 \, \text{J/mol}$

Answer with Three Significant Figures

$\Delta G = -2.12 \times 10^5 \, \text{J/mol}$

Thus, the theoretical $\Delta G$ for the copper-zinc voltaic cell is:

$\Delta G = -2.12 \times 10^5 \, \text{J/mol}$

Would you like additional details on any step, or do you have further questions?

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Here are some related questions:

1. How would changes in concentration affect the $\Delta G$ of this reaction?
2. What is the role of the salt bridge in a copper-zinc voltaic cell?
3. How does the $E_{\text{cell}}$ change if the reaction is not under standard conditions?
4. Can you explain why the zinc electrode is the anode in this cell?
5. How does the number of electrons transferred ($n$) affect the $\Delta G$ calculation?

Tip: Always remember to check the sign of $\Delta G$ to confirm whether a reaction is spontaneous (negative $\Delta G$) or non-spontaneous (positive $\Delta G$).