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: Link work to cell potential:
Potential is work per unit charge, so work equals charge times potential.
For n moles of electrons the charge passed is $nF$ coulombs.

Step 2: Link work to Gibbs energy:
At constant temperature and pressure, the maximum useful work done by a system is $-\Delta G$.
So $-\Delta_rG^{\ominus} = nF\,E^{\ominus}_{cell}$. Rearranging gives $\Delta_rG^{\ominus} = -nF\,E^{\ominus}_{cell}$.

Step 3: Sanity check with a spontaneous cell:
For the Daniell cell, $E^{\ominus} = 1.10$ V and $n = 2$.
\[ \Delta_rG^{\ominus} = -2 \times 96500 \times 1.10 = -212300\ \mathrm{J\ mol^{-1}} \approx -212\ \mathrm{kJ\ mol^{-1}} \]
The value is negative, as a spontaneous reaction needs. This confirms the minus sign.

Step 4: Rule out the rest:
(A) drops $nF$, and a volt cannot equal a joule per mole.
(B) has the positive sign, which would make every working cell non-spontaneous.
(C) lacks $F$, so its units are wrong.

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