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For the cell reaction below, what is the standard Gibbs free energy change at 298 K? Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Given that E° = 2.0 V at 298 K and Faraday’s constant F = 96000 C mol−1, find ΔG° in kJ mol−1.

1. −384
2. 384
3. 192
4. −192

Correct answer: −384

Solution

The standard Gibbs free energy change (ΔG°) is calculated using the equation ΔG° = -nFE°, where n is the number of moles of electrons transferred in the reaction. In this case, 2 moles of electrons are transferred, resulting in ΔG° = -2 * 96000 C/mol * 2.0 V = -384 kJ/mol, confirming that option A is correct.

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