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: Understanding the Concept:
A galvanic cell turns chemical energy into electrical work.
The maximum useful work from the cell equals the decrease in Gibbs energy.

Step 2: Key Formula or Approach:
Electrical work is charge times potential difference.
The charge passed for n moles of electrons is \( nF \), where F = 96500 C per mole.
The potential difference at standard conditions is \( \mathrm{E^{\ominus}_{cell}} \).

Step 3: Detailed Explanation:
The maximum electrical work done by the cell is \( nF\,\mathrm{E^{\ominus}_{cell}} \).
Work done by the system is taken as a loss of Gibbs energy.
So the change in Gibbs energy carries a negative sign:
\[ \Delta_r G^{\ominus} = -\, nF\, \mathrm{E^{\ominus}_{cell}} \]
A positive cell potential gives a negative \( \Delta_r G^{\ominus} \), so the reaction is spontaneous.
Option (A) has no n and no F, and it has the wrong sign.
Option (B) has the right terms but the wrong sign.
Option (C) has no F, so its units are wrong.
Option (D) is the correct relation.

Step 4: Final Answer:
The correct relation is option (D). \[ \boxed{\Delta_r G^{\ominus} = -\, nF\, \mathrm{E^{\ominus}_{cell}}} \]