Question

In the reaction, \( \text{O}_{2(g)} + 4\text{H}^+_{(aq)} + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}_{(l)} \), the quantity of electricity required to reduce one mole of gaseous oxygen is ( \( E^\circ = 1.23\,\text{V} \) )

Hint:

Always read the stoichiometry directly from the balanced equation. If a question asks for charge, you don't need voltage ($E^\circ$). Voltage is needed to calculate energy ($\Delta G = -nFE$). Don't let extra numbers distract you.

Answer 1

Step 1: Identify Electron Requirement
From the balanced reaction: \[ \text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O} \] 1 mole of $\text{O}_2$ requires 4 moles of electrons.

Step 2: Use Faraday's Law
\[ Q = nF \] where $n =$ number of moles of electrons, $F = 96500 \text{ C/mol}$

Step 3: Substitute Values
\[ Q = 4 \times 96500 \]

Step 4: Calculate Charge
\[ Q = 386000 \text{ C} \]

Step 5: Express in Faradays
\[ Q = 4F \]

Step 6: Final Answer
\[ \boxed{Q = 4F = 3.86 \times 10^5 \text{ C}} \]