Question

The relationship between \( E^{o}_{(cell)} \) and \( \Delta_rG^{\ominus} \) is :

• A. \( \Delta_rG^{\ominus} = E^{\ominus}_{(cell)} \)
• B. \( \Delta_rG^{\ominus} = nF\,E^{\ominus}_{(cell)} \)
• C. \( \Delta_rG^{\ominus} = nE^{o}_{cell} \)
• D. \( \Delta_rG^{\ominus} = -\,nF\,E^{\ominus}_{(cell)} \)

Hint:

Electrical work equals nFE and it comes at the expense of Gibbs energy, so delta G = -nFE.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A galvanic cell does electrical work when it runs. The maximum work equals the decrease in Gibbs energy.
So the cell potential and the Gibbs energy of the cell reaction are linked.

Step 2: Key Formula or Approach:
The electrical work for n moles of electrons is the charge nF times the potential E.
Work done by the system lowers the Gibbs energy, so a minus sign appears.
\[ \Delta_rG^{\ominus} = -\,nF\,E^{\ominus}_{cell} \]
Here n is the number of electrons exchanged and F is the Faraday constant, 96500 C per mole.

Step 3: Detailed Explanation:
Option (A) has no n, no F and no minus sign. Units of energy cannot equal units of volt, so (A) is wrong.
Option (B) has the right size but the wrong sign. A spontaneous cell has positive E, so \( \Delta_rG^{\ominus} \) must be negative. This option would give a positive value, so (B) is wrong.
Option (C) has no Faraday constant. Without F the units are not those of energy. So (C) is wrong.
Option (D) has the minus sign and nF, so it is correct.

Final Answer:
The relation is \( \Delta_rG^{\ominus} = -nFE^{\ominus}_{cell} \), which is option (D).\[ \boxed{\Delta_rG^{\ominus} = -\,nF\,E^{\ominus}_{cell}} \]