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 Cl2 gas at STP are produced?

Question

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 [latex]Cl_{2}[/latex] 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.

Accepted Answer

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:

[latex]Charge =(30.0 \frac{C}{s})(1.00 h) (\frac{60 min}{1 h})(\frac{60 s}{1 min})=1.08 × 10^{5} C[/latex]

Moles of [latex]e^{-} = (1.08 × 10^{5} C)(\frac{1 mol e^{-}}{96,500 C})=1.12 mol e^{-}[/latex]

The cathode reaction yields 2 mol of [latex]OH^{-}[/latex] per 2 mol of electrons (Section 19.12: Electrolysis of Aqueous Sodium Chloride), so 1.12 mol of NaOH will be obtained:

[latex]2 H_{2}O(l) + 2 e^{-} → H_{2}(g) + 2 OH^{-}(aq)[/latex]

Moles of [latex]NaOH = (1.12 mol e^{-})(\frac{2 mol NaOH}{2 mol e^{-}})=1.12 mol NaOH[/latex]

Converting the number of moles of NaOH to grams gives 44.8 g of NaOH:

Grams of [latex]NaOH = (1.12 mol NaOH)(\frac{40.0 g NaOH}{1 mol NaOH})=44.8 g NaOH[/latex]

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

[latex](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})[/latex]= 44.8 g NaOH

The anode reaction gives 1 mol of [latex]Cl_{2}[/latex] per 2 mol of electrons, so 0.560 mol of [latex]Cl_{2}[/latex] will be obtained:

[latex]2 Cl^{-}(aq) → Cl_{2}(g) + 2 e^{-}[/latex]

Moles of [latex]Cl_{2} = (1.12 mol e^{-})(\frac{1 mol Cl_{2}}{2 mol e^{-} })=0.560 mol Cl_{2}[/latex]

Since 1 mol of an ideal gas occupies 22.4 L at STP, the volume of [latex]Cl_{2}[/latex] obtained is 12.5 L:

Liters of [latex]Cl_{2} = (0.560 mol Cl_{2})(\frac{22.4 L Cl_{2}}{mol Cl_{2}})= 12.5 L Cl_{2}[/latex]

As a shortcut, the entire sequence of conversions can be carried out in one step. For example, the volume of [latex]Cl_{2}[/latex] produced at the anode is

[latex](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}})[/latex][latex]= 12.5 L Cl_{2}[/latex]

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 [latex]Cl_{2}[/latex] per mole of electrons, 1 mol of NaOH (∼40 g) and 0.5 mol of [latex]Cl_{2}[/latex] (∼11 L at STP) will be formed. The estimate and the solution agree.

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)

Question 19.13: calculating the amount of product produced by Electrolysis A...

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