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Difficulty: MediumFaraday's Laws of Electrolysis and Quantitative Calculations

Calculate the volume of oxygen gas, in cm3\text{cm}^3 measured at STP, liberated at the anode during the electrolysis of dilute tetraoxosulfate(VI) acid when a steady current of 1.93 A1.93\text{ A} is passed through the electrolyte for 50 minutes50\text{ minutes}.

[Take Faraday's constant F=96500 C mol−1F = 96500\text{ C mol}^{-1}, Molar volume of gas at STP =22.4 dm3 mol−1= 22.4\text{ dm}^3\text{ mol}^{-1}]

Answer: 336 cm³

Answer

The volume of oxygen gas liberated at STP is 336 cm3336\text{ cm}^3.
Passing a current of 1.93 A1.93\text{ A} for 3000 s3000\text{ s} delivers 5790 C5790\text{ C} of charge, equivalent to 0.06 moles0.06\text{ moles} of electrons. Because the anodic discharge of hydroxide ions (4OH−→2H2O+O2+4e−4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-) requires 4 moles4\text{ moles} of electrons per mole of O2\text{O}_2, 0.015 moles0.015\text{ moles} of O2\text{O}_2 are generated. Multiplying by the molar volume at STP (22400 cm3mol−122400\text{ cm}^3\text{mol}^{-1}) gives 336 cm3336\text{ cm}^3.

Step-by-Step Solution

1
Convert the duration of electrolysis from minutes to seconds
t=50 min×60 s/min=3000 st = 50\text{ min} \times 60\text{ s/min} = 3000\text{ s}
Current calculations require time in SI units (seconds).
2
Calculate the total electric charge passed through the electrolyte
Q=I×t=1.93 A×3000 s=5790 CQ = I \times t = 1.93\text{ A} \times 3000\text{ s} = 5790\text{ C}
Charge passed is the product of electric current and time.
3
Calculate the quantity of electrons passed in moles
n(e−)=5790 C96500 C mol−1=0.06 moln(e^-) = \frac{5790\text{ C}}{96500\text{ C mol}^{-1}} = 0.06\text{ mol}
One mole of electrons corresponds to 1 Faraday (96500 C96500\text{ C}).
4
Use the anodic half-reaction equation to determine the molar ratio of electrons to oxygen gas
4OH−(aq)→2H2O(l)+O2(g)+4e−4\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g) + 4e^-; n(O2)=0.06 mol4=0.015 moln(\text{O}_2) = \frac{0.06\text{ mol}}{4} = 0.015\text{ mol}
The discharge of hydroxide ions requires 4 moles of electrons per mole of oxygen gas evolved.
5
Calculate the volume of liberated oxygen gas at STP in cm3\text{cm}^3
V=0.015 mol×22400 cm3 mol−1=336 cm3V = 0.015\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 336\text{ cm}^3
One mole of gas occupies 22.4 dm3=22400 cm322.4\text{ dm}^3 = 22400\text{ cm}^3 at standard temperature and pressure.

Key Concept

Quantitative electrochemistry using Faraday's laws and stoichiometric electron-to-gas relationships at electrodes.
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Topics

Faraday's Laws of Electrolysis and Quantitative Calculations

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