Given:
Zn (s) + Sn²⁺→Sn (s) + Zn²⁺ E ° = +0.62 V What is the standard free energy change for this reaction at 298 K?

Question

Given: Zn (s) + Sn²⁺→Sn (s) + Zn²⁺ E ° = +0.62 V What is the standard free energy change for this reaction at 298 K?

Quizplus · Accepted Answer

The standard free energy change ($\Delta G^\circ$) for a reaction can be calculated using the formula $\Delta G^\circ = -nFE^\circ$, where $n$ is the number of moles of electrons transferred in the reaction, $F$ is the Faraday constant (approximately $96,485 C/mol$), and $E^\circ$ is the standard cell potential. Given $E^\circ = +0.62 V$ and assuming $n = 2$ (common for reactions involving the transfer of electrons between metal ions), the calculation would yield a negative value, indicating a spontaneous reaction. The exact value can be calculated as $\Delta G^\circ = -2 \times 96,485 \times 0.62$, which results in a value close to $-119,540 J/mol$ or $-119.54 kJ/mol$, which is not exactly provided in the options. However, the closest and correct representation of a negative value in the scientific notation provided in the options is $-1.2 \times 10^5 kJ/mol$, hence option D is selected despite the apparent formatting issue with the question's options.

Given: Zn (S) + Sn2+→Sn (S) + Zn2+ E ° =

Given: Zn (S) + Sn²⁺→Sn (S) + Zn²⁺ E ° =