Question 19.8: A current of 1.26 A is passed through an electrolytic cell c...
A current of 1.26 \text{A} is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 \text{h}. Write the half-cell reactions, and calculate the volume of gases generated at \text{STP}.
Strategy As shown in Figure 19.12, we can use current and time to determine charge. We can then convert charge to moles of electrons, and use the balanced half-reactions to determine how many moles of product form at each electrode. Finally, we can convert moles to volume.
Setup The half-cell reactions for the electrolysis of water are
Anode: 2\text{H}_2\text{O}(l) → \text{O}_2(\text{g}) + 4\text{H}^+(aq) + 4e^–
Cathode: \underline{4\text{H}^+(aq) + 4e^– → 2\text{H}_2(\text{g}) }
Overall: 2\text{H}_2\text{O}(l) → \text{O}_2(\text{g}) + 2\text{H}_2(\text{g})
Remember that \text{STP} for gases means 273 \text{K} and 1 \text{atm}
\text{coulombs} = (1.26 \cancel{\text{A}})(7.44 \cancel{\text{h}}) \left(\frac{3600 \cancel{\text{s}}}{1 \cancel{\text{h}}} \right) \left(\frac{1 \text{C}}{1 \cancel{\text{A ⋅ s}}}\right) = 3.375 × 10^4 \text{C}
At the anode:
\text{moles O}_2 = (3.375 × 10^4 \cancel{\text{C}}) \left(\frac{1 \cancel{\text{mol} e}} {96,500 \cancel{\text{C}}} \right) \left(\frac{1 \text{mol O}_2}{4 \cancel{\text{mol} e}} \right) = 0.0874 \text{mol O}_2
The volume of 0.0874 \text{mol O}_2 at \text{STP} is given by
V = \frac{nRT}{P}
= \frac{(0.0874 \text{mol})(0.08206 \text{L ⋅ atm/K ⋅ mol})(273.15 \text{K})}{1 \text{atm}} = 1.96 \text{L O}_2
Similarly, for hydrogen we write
\text{moles H}_2 = (3.375 × 10^4 \cancel{\text{C}}) \left(\frac{1 \cancel{\text{mol} e}} {96,500 \cancel{\text{C}}} \right) \left(\frac{1 \text{mol H}_2}{2 \cancel{\text{mol} e}} \right) = 0.175 \text{mol H}_2
The volume of 0.175 \text{mol H}_2 at \text{STP} is given by
V = \frac{nRT}{P}
= \frac{(0.175 \text{mol})(0.08206 \text{L ⋅ atm/K ⋅ mol})(273.15 \text{K})}{1 \text{atm}} = 3.92 \text{L H}_2