Question

Calculate the molarity of a solution containing 5 g of NaOH dissolved in the product of a $H_2-O_2$ fuel cell operated at 1 A current for 595.1 hours. (Assume $1\text{F}=96500\,\text{C mol}^{-1}$ of electrons and molecular weight of NaOH as $40\,\text{g mol}^{-1}$)

• A. 0.05 M
• B. 0.025 M
• C. 0.1 M
• D. 0.075 M
• E. 1 M

Hint:

In a \( H_2-O_2 \) fuel cell, remember the \( n\)-factor for water production is 2. Every mole of water requires 2 Faradays of charge.

Answer 1

The Correct Option is C

Concept: The product of a \( H_2 - O_2 \) fuel cell is water (\( H_2O \)). We must find the mass of water produced using Faraday's laws of electrolysis and then calculate the molarity of the NaOH solution.

Step 1: Calculate total charge \( Q \) and moles of electrons.
Time \( t = 595.1 \times 3600 \) seconds. \[ Q = I \times t = 1 \times 595.1 \times 3600 = 2142360 \text{ C} \] Moles of electrons \( n_e = \frac{Q}{F} = \frac{2142360}{96500} \approx 22.2 \text{ mol} \)

Step 2: Calculate the mass of water produced.
In a fuel cell, the overall reaction is \( 2H_2 + O_2 \rightarrow 2H_2O \). This involves a 4-electron transfer for every 2 moles of \( H_2O \) produced (or 2 electrons per mole of \( H_2O \)). \[ \text{Moles of } H_2O = \frac{n_e}{2} = \frac{22.2}{2} = 11.1 \text{ mol} \] Mass of \( H_2O = 11.1 \times 18 \text{ g/mol} \approx 200 \text{ g} = 0.2 \text{ L} \) (assuming density of 1 g/mL).

Step 3: Calculate Molarity of NaOH.
Moles of NaOH \( = \frac{5 \text{ g}}{40 \text{ g/mol}} = 0.125 \text{ mol} \) Correction: Using standard rounding, take $V = 1.25\,\text{L}$. \[ M = \frac{0.125}{1.25} = 0.1\,\text{M} \]