Question

Consider a fuel cell supplied with 1 mole of H$_2$ gas and 10 moles of O$_2$ gas. If the fuel cell is operated at 96.5 mA current, how long will it deliver power? (Assume $1\ F = 96500\ \text{C/mol of electrons}$)

• A. $1 \times 10^6$ s
• B. $0.5 \times 10^6$ s
• C. $2 \times 10^6$ s
• D. $4 \times 10^6$ s
• E. $5 \times 10^6$ s

Hint:

Always find limiting reagent first, then electrons transferred.

Answer 1

The Correct Option is C

Concept:
In a fuel cell: \[ \text{H}_2 \rightarrow 2H^+ + 2e^- \] 1 mole of H$_2$ releases 2 moles of electrons.
Total charge: \[ Q = nF \] Time: \[ t = \frac{Q}{I} \]

Step 1: Identify limiting reactant.
Reaction: \[ 2H_2 + O_2 \rightarrow 2H_2O \] 1 mole H$_2$ requires 0.5 mole O$_2$. Since O$_2$ is excess, H$_2$ is limiting.

Step 2: Total electrons produced.
\[ 1\ \text{mol H}_2 \Rightarrow 2\ \text{mol e}^- \]

Step 3: Total charge.
\[ Q = 2 \times 96500 = 193000\ \text{C} \]

Step 4: Convert current.
\[ I = 96.5\ \text{mA} = 0.0965\ \text{A} \]

Step 5: Time calculation.
\[ t = \frac{193000}{0.0965} = 2 \times 10^6\ \text{s} \]