Q.In a fuel cell, $\text{H}_2$ and $\text{O}_2$ react to produce electricity. In the process, $\text{H}_2$ gas is oxidised at the anode and $\text{O}_2$ gas is reduced at the cathode. If 44.8 litres of $\text{H}_2$ at 25°C and 1 atm pressure reacts in 10 minutes, what is the average current produced? If the entire current is used for electrodeposition of Cu from $\text{Cu}^{2+}$, how many grams of Cu are deposited?

Step 1. Moles of $\text{H}_2$ reacted: using the standard molar volume convention (22.4 L/mol), $n(\text{H}_2) = 44.8/22.4 = 2$ mol.

Step 2. At the anode, $\text{H}_2 \rightarrow 2\text{H}^++2e^-$ (in the acidic-medium form; equivalently the alkaline form releases the same 2 electrons per H₂), so 2 mol $\text{H}_2$ releases $2\times2 = 4$ mol electrons.

Step 3. Charge: $Q = 4\times96500 = 386000$ C. Time: $t = 10\times60=600$ s. Average current: $I = Q/t = 386000/600 \approx 643.3$ A. …

Convert the gas volume to moles (using 22.4 L/mol), find the electrons released via the anode half-reaction, compute Q and then I= …

Using the wrong n-factor for hydrogen (it is 2 electrons per H2 molecule, not 1), or forgetting that the SAME charge (not current) ca …