Question

The relationship between \( \mathrm{E^{\circ}_{(cell)}} \) and \( \Delta_r G^{\ominus} \) is :

• A. \( \Delta_r G^{\ominus} = \mathrm{E^{\ominus}_{(cell)}} \)
• B. \( \Delta_r G^{\ominus} = nF\, \mathrm{E^{\ominus}_{(cell)}} \)
• C. \( \Delta_r G^{\ominus} = n\, \mathrm{E^{\circ}_{cell}} \)
• D. \( \Delta_r G^{\ominus} = -\, nF\, \mathrm{E^{\ominus}_{(cell)}} \)

Hint:

Maximum electrical work equals nF times the cell potential and it equals the fall in Gibbs energy, so a minus sign appears.

Answer 1

The Correct Option is D

Step 1: Check the units first:
\( \Delta_r G^{\ominus} \) is in joule per mole.
The product of charge and potential gives coulomb times volt, which is joule.
So the formula must have the charge \( nF \) multiplying the cell potential.
This rules out option (A), which has no charge, and option (C), which has no F.

Step 2: Decide the sign:
A cell with positive \( \mathrm{E^{\ominus}_{cell}} \) works on its own, so the reaction is spontaneous.
A spontaneous reaction has a negative \( \Delta_r G^{\ominus} \).
Option (B) has a plus sign, so it would make a spontaneous cell non-spontaneous.
That removes (B).

Step 3: Put it together:
Only option (D) has the right terms and the right sign:
\[ \Delta_r G^{\ominus} = -\, nF\, E^{\ominus}_{cell} \]
Check with the Daniell cell: n = 2 and E = 1.1 V give \( \Delta_r G^{\ominus} = -2 \times 96500 \times 1.1 \approx -212 \) kJ per mole, which is negative as expected.

Final Answer:
Option (D) is correct. \[ \boxed{\Delta_r G^{\ominus} = -\, nF\, E^{\ominus}_{cell}} \]