Question 19.13: calculating the amount of product produced by Electrolysis A...
Calculating the Amount of Product Produced by Electrolysis
A constant current of 30.0 A is passed through an aqueous solution of NaCl for 1.00 h. How many grams of NaOH and how many liters of Cl_{2} gas at STP are produced?
STRATEGY
To convert the current and time to grams or liters of product, carry out the sequence of conversions in Figure 19.18. Another strategy is to start with the known unit and use conversions to reach the desired unit.
IDENTIFY
| Known | Unknown |
| NaCl (aq) | Amount of products (g NaOH and L of Cl_{2}) |
| Time (1.00 h) | |
| Current (30.0 A = 30.0 C/s) | |
| STP (at STP 1 mol of gas has a volume of 22.4 L) | |
Because electrons can be thought of as a reactant in the electrolysis process, the first step is to calculate the charge and the number of moles of electrons passed through the cell:
Charge =(30.0 \frac{C}{s})(1.00 h) (\frac{60 min}{1 h})(\frac{60 s}{1 min})=1.08 × 10^{5} CMoles of e^{-} = (1.08 × 10^{5} C)(\frac{1 mol e^{-}}{96,500 C})=1.12 mol e^{-}
The cathode reaction yields 2 mol of OH^{-} per 2 mol of electrons (Section 19.12: Electrolysis of Aqueous Sodium Chloride), so 1.12 mol of NaOH will be obtained:
2 H_{2}O(l) + 2 e^{-} → H_{2}(g) + 2 OH^{-}(aq)Moles of NaOH = (1.12 mol e^{-})(\frac{2 mol NaOH}{2 mol e^{-}})=1.12 mol NaOH
Converting the number of moles of NaOH to grams gives 44.8 g of NaOH:
Grams of NaOH = (1.12 mol NaOH)(\frac{40.0 g NaOH}{1 mol NaOH})=44.8 g NaOH
As a shortcut, the entire sequence of conversions can be carried out in one step. For example, the mass of NaOH produced at the cathode is
(30.0 \frac{C}{s})(1.00 h)(\frac{3600 s}{h})(\frac{1 mol e^{-}}{96,500 C})(\frac{2 mol NaOH}{2 mol e^{-}})(\frac{40.0 g NaOH}{1 mol NaOH})= 44.8 g NaOH
The anode reaction gives 1 mol of Cl_{2} per 2 mol of electrons, so 0.560 mol of Cl_{2} will be obtained:
2 Cl^{-}(aq) → Cl_{2}(g) + 2 e^{-}Moles of Cl_{2} = (1.12 mol e^{-})(\frac{1 mol Cl_{2}}{2 mol e^{-} })=0.560 mol Cl_{2}
Since 1 mol of an ideal gas occupies 22.4 L at STP, the volume of Cl_{2} obtained is 12.5 L:
Liters of Cl_{2} = (0.560 mol Cl_{2})(\frac{22.4 L Cl_{2}}{mol Cl_{2}})= 12.5 L Cl_{2}
As a shortcut, the entire sequence of conversions can be carried out in one step. For example, the volume of Cl_{2} produced at the anode is
(30.0\frac{C}{s})(1.00 h)(\frac{3600 s}{h})(\frac{1 mol e^{-}}{96,500 C})(\frac{1 mol Cl_{2}}{2 mol e^{-}})(\frac{22.4 L Cl_{2}}{mol Cl_{2}})= 12.5 L Cl_{2}
CHECK
Since approximately 1 mol of electrons is passed through the cell and the electrode reactions yield 1 mol of NaOH and 0.5 mol of Cl_{2} per mole of electrons, 1 mol of NaOH (∼40 g) and 0.5 mol of Cl_{2} (∼11 L at STP) will be formed. The estimate and the solution agree.