[VISUAL] Based on the information in the table above, which of the following shows the cell potentia — Electrochemistry Chemistry Question

STEPS:

1. Identify the oxidation and reduction half-reactions for a galvanic cell:
* A standard galvanic cell must have a positive standard cell potential ($E^\circ_{\text{cell}} > 0$) because the reaction is thermodynamically favored.
* The two standard reduction potentials given in the table are:
* $\text{Mg}^{2+}(aq) + 2e^- \rightarrow \text{Mg}(s) \quad E^\circ_{\text{red}} = -2.37 \text{ V}$
* $\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ_{\text{red}} = -0.74 \text{ V}$
* To achieve a positive overall cell potential, the half-reaction with the more negative reduction potential ($\text{Mg}^{2+}/\text{Mg}$) must be reversed to act as the oxidation half-reaction at the anode:
* Anode (Oxidation): $\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^- \quad E^\circ_{\text{ox}} = +2.37 \text{ V}$
* Cathode (Reduction): $\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ_{\text{red}} = -0.74 \text{ V}$

2. Calculate the standard cell potential ($E^\circ_{\text{cell}}$):
* Use the standard cell potential formula:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
$$E^\circ_{\text{cell}} = -0.74 \text{ V} - (-2.37 \text{ V}) = \mathbf{+1.63 \text{ V}}$$
* *Note: Standard reduction potential is an intensive property and does not change when half-reactions are multiplied by stoichiometric coefficients.*

3. Determine the number of moles of electrons transferred ($n$):
* To balance the electrons transferred in the overall redox reaction, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
* $3 \times (\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-)$ (6 moles of electrons lost)
* $2 \times (\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s))$ (6 moles of electrons gained)
* Overall balanced equation: $3\text{Mg}(s) + 2\text{Cr}^{3+}(aq) \rightarrow 3\text{Mg}^{2+}(aq) + 2\text{Cr}(s)$
* Thus, $n = 6 \text{ moles of electrons}$ transferred per mole of reaction.

4. Calculate the standard Gibbs free energy change ($\Delta G^\circ$):
* Use the thermodynamic relationship:
$$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$
* Substitute the values ($n = 6$, $F \approx 96,485 \text{ C/mol } e^-$, and $E^\circ_{\text{cell}} = 1.63 \text{ V} = 1.63 \text{ J/C}$):
$$\Delta G^\circ = -(6 \text{ mol } e^-)\left(96,485 \frac{\text{C}}{\text{mol } e^-}\right)\left(1.63 \frac{\text{J}}{\text{C}}\right)$$
$$\Delta G^\circ \approx -943,623 \text{ J/mol}_{\text{rxn}} = \mathbf{-944 \text{ kJ/mol}_{\text{rxn}}}$$

5. Select the correct option:
* A standard cell potential of $+1.63 \text{ V}$ combined with a $\Delta G^\circ$ of $-944 \text{ kJ/mol}_{\text{rxn}}$ corresponds to Option B.

*

WHY_OTHERS_WRONG:

* Option A is incorrect: This option has the correct cell potential ($+1.63 \text{ V}$) but calculates $\Delta G^\circ$ incorrectly by using $n = 1$ rather than $n = 6$ in the free energy formula:
$$\Delta G^\circ = -(1)(96,485 \text{ C/mol})(1.63 \text{ V}) \approx -157 \text{ kJ/mol}_{\text{rxn}}$$
* Option C is incorrect: This option makes the fundamental error of treating standard reduction potential as an extensive property by multiplying the half-reaction potentials by their stoichiometric coefficients ($3 \times 2.37 \text{ V} = 7.11 \text{ V}$ and $2 \times -0.74 \text{ V} = -1.48 \text{ V}$), resulting in an incorrect cell potential of $+5.63 \text{ V}$. It then also incorrectly uses $n = 1$ to calculate the $\Delta G^\circ$ value of $-543 \text{ kJ/mol}_{\text{rxn}}$.
* Option D is incorrect: This option combines the incorrect extensive-property potential calculation ($E^\circ_{\text{cell}} = +5.63 \text{ V}$) with the correct $n = 6$ electron transfer to yield a massive, incorrect free energy change of $-3262 \text{ kJ/mol}_{\text{rxn}}$.