Question

The standard free energy change for the following reaction is -210 kJ. What is the standard cell potential (in V)?
$2H_2O(aq) \rightarrow 2H_2(g) + O_2(g)$
HINT: $\Delta G^\circ = -n F E^\circ$

• A. +0.752
• B. +1.09
• C. +0.544
• D. +0.640

Answer 1

Answer: (C)

+0.544 V

Answer 2

Step 1: Identify $\Delta G^\circ$ and relate to $E^\circ$.
$\Delta G^\circ = -nFE^\circ$
Given $\Delta G^\circ = -210\text{ kJ} = -210000\text{ J}$.

Step 2: Determine number of electrons ($n$).
Reaction:
$\displaystyle 2H_2O \to 2H_2 + O_2$
This corresponds to transfer of 4 electrons (combining the standard H₂/H⁺ and O₂/H₂O half‐reactions).

Step 3: Use Faraday’s constant $F = 96485\text{ C/mol}$.

$$-210000 = -\,n\,F\,E^\circ \quad\Longrightarrow\quad E^\circ = \frac{210000}{n\,F} = \frac{210000}{4 \times 96485} \approx 0.544\text{ V}$$

Answer: $E^\circ = +0.544\text{ V}$.