overall reaction for the hydrogen-oxygen fuel cell is:

$ 2 \mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O} $

From this reaction, we see that 4 Faradays (F) of charge will produce 2 moles of water. One Faraday is approximately 96,500 Coulombs (C), so 4 F equals 4 × 96,500 C.

Calculating Charge Passed:

The charge ($ Q $) is given by the current ($ I $) multiplied by the time ($ t $). Here, the current is 1 A, and the time is 595.1 hours. Convert this time into seconds:

$ t = 595.1 \times 60 \times 60 \, \text{seconds} $

$ Q = 1 \times 595.1 \times 60 \times 60 = 2,142,360 \, \text{C} $

Determine Water Produced:

Use the relationship between charge available and water produced:

$ \text{Water produced} = \frac{36 \, \text{g}}{4 \times 96,500 \, \text{C}} \times 2,142,360 \, \text{C} \approx 200 \, \text{g} $

This is equivalent to 200 mL of water because the density of water is 1 g/mL.

Calculate Molarity of NaOH Solution:

The NaOH is dissolved in this water. We know:

Mass of NaOH = 5 g

Molar mass of NaOH = 40 g/mol

Volume of water in liters = 200 mL = 0.200 L

Calculate moles of NaOH:

$ \text{Moles of NaOH} = \frac{5 \, \text{g}}{40 \, \text{g/mol}} = 0.125 \, \text{mol} $

Finally, calculate molarity:

$ \text{Molarity} = \frac{\text{moles of NaOH}}{\text{volume in liters}} = \frac{0.125 \, \text{mol}}{0.200 \, \text{L}} = 0.625 \, \text{M} $

Thus, the molarity of the NaOH solution is 0.625 M.

Electrochemistry

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