The standard reduction potentials for the half-reactions related to the galvanic cell represented ab — Electrochemistry Chemistry Question

STEPS:

1. Understand the thermodynamic nature of a galvanic cell: A galvanic (or voltaic) cell is an electrochemical cell that converts chemical energy into electrical energy using a thermodynamically favored (spontaneous) redox reaction. For any galvanic cell to operate spontaneously, its overall standard cell potential ($E^\circ_{\text{cell}}$) must be positive ($E^\circ_{\text{cell}} > 0$).
2. Determine which species undergoes reduction and which undergoes oxidation:
* Standard reduction potentials ($E^\circ$) measure a species' tendency to gain electrons and be reduced. The half-reaction with the more positive (or less negative) $E^\circ$ value represents the stronger oxidizing agent, which will proceed as a reduction at the cathode.
* Comparing the two given reduction potentials:
* $\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ = -0.74\text{ V}$
* $\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\circ = -0.76\text{ V}$
* Because $-0.74\text{ V}$ is more positive than $-0.76\text{ V}$, $\text{Cr}^{3+}$ has a stronger pull for electrons. Consequently, $\text{Cr}^{3+}$ will be reduced at the cathode, and solid zinc ($\text{Zn}$) must be oxidized at the anode.
3. Identify the half-reactions at each electrode:
* Cathode (Reduction): $\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ_{\text{cathode}} = -0.74\text{ V}$
* Anode (Oxidation): $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \quad E^\circ_{\text{anode}} = -0.76\text{ V}$
4. Calculate the standard cell potential ($E^\circ_{\text{cell}}$):
* Use the standard cell potential formula:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
*(where both potentials on the right-hand side of the equation are standard reduction potentials)*
* Substitute the values from the table:
$$E^\circ_{\text{cell}} = (-0.74\text{ V}) - (-0.76\text{ V})$$
$$E^\circ_{\text{cell}} = -0.74\text{ V} + 0.76\text{ V} = \mathbf{+0.02\text{ V}}$$
* *Note on stoichiometry:* Standard reduction potentials are intensive properties. Even though you would multiply the chromium half-reaction by 2 and the zinc half-reaction by 3 to balance the overall electron transfer (6 electrons), you do not multiply their $E^\circ$ values when calculating the cell potential.
5. Select the correct option: The calculated cell potential is $+0.02\text{ V}$, confirming Option D is the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect ($E_{\text{cell}} = -1.50\text{ V}$): This value is calculated by adding the two reduction potentials directly together without reversing the zinc half-reaction and keeping both signs negative ($-0.74\text{ V} + (-0.76\text{ V}) = -1.50\text{ V}$). Additionally, a negative cell potential indicates a non-spontaneous process, which cannot occur in an operating galvanic cell.
• Option B is incorrect ($E_{\text{cell}} = -0.80\text{ V}$): This is mathematically incorrect and represents a non-spontaneous cell. A student might arrive at this number due to miscellaneous sign or arithmetic errors.
• Option C is incorrect ($E_{\text{cell}} = -0.02\text{ V}$): This is the negative of the correct value. A student would calculate this if they incorrectly designated zinc as the cathode and chromium as the anode ($E^\circ_{\text{cell}} = E^\circ_{\text{Zn}} - E^\circ_{\text{Cr}} = -0.76\text{ V} - (-0.74\text{ V}) = -0.02\text{ V}$). Because galvanic cells must have a positive potential to run spontaneously, a negative value is physically impossible for this operating system.