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Question: In electrolysis of a concentrated solution of $H_2SO_4$ to form peroxydisulphuric acid ($H_2S_2O_8$)...

In electrolysis of a concentrated solution of H2SO4H_2SO_4 to form peroxydisulphuric acid (H2S2O8H_2S_2O_8), O2O_2 and H2H_2 are produced as by products on respective electrodes. If 2.8 litre of O2O_2 and 11.2 litre of H2H_2 is produced at STP. The weight of H2S2O8H_2S_2O_8 formed will be

Answer

48.5 g

Explanation

Solution

To solve this problem, we need to consider the electrochemical reactions occurring at the anode and cathode and apply Faraday's laws of electrolysis.

1. Calculate moles of gases produced: At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Moles of O2=Volume of O2Molar volume at STP=2.8 L22.4 L/mol=0.125 molO_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{2.8 \text{ L}}{22.4 \text{ L/mol}} = 0.125 \text{ mol} Moles of H2=Volume of H2Molar volume at STP=11.2 L22.4 L/mol=0.5 molH_2 = \frac{\text{Volume of } H_2}{\text{Molar volume at STP}} = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ mol}

2. Write down the electrode reactions and associated electron transfer: At the anode (oxidation): a) Formation of peroxydisulphuric acid (H2S2O8H_2S_2O_8): 2H2SO4→H2S2O8+2H++2e−2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^- For 1 mole of H2S2O8H_2S_2O_8, 2 moles of electrons are transferred.

b) Formation of oxygen gas (O2O_2): 2H2O→O2+4H++4e−2H_2O \rightarrow O_2 + 4H^+ + 4e^- For 1 mole of O2O_2, 4 moles of electrons are transferred.

At the cathode (reduction): a) Formation of hydrogen gas (H2H_2): 2H++2e−→H22H^+ + 2e^- \rightarrow H_2 For 1 mole of H2H_2, 2 moles of electrons are transferred.

3. Relate moles of products to moles of electrons: Let nH2S2O8n_{H_2S_2O_8} be the moles of H2S2O8H_2S_2O_8 formed. Let nO2n_{O_2} be the moles of O2O_2 formed. Let nH2n_{H_2} be the moles of H2H_2 formed.

Moles of electrons involved in H2S2O8H_2S_2O_8 formation = 2×nH2S2O82 \times n_{H_2S_2O_8} Moles of electrons involved in O2O_2 formation = 4×nO2=4×0.125 mol=0.5 mol4 \times n_{O_2} = 4 \times 0.125 \text{ mol} = 0.5 \text{ mol} Moles of electrons involved in H2H_2 formation = 2×nH2=2×0.5 mol=1.0 mol2 \times n_{H_2} = 2 \times 0.5 \text{ mol} = 1.0 \text{ mol}

4. Apply Faraday's Law (conservation of charge): The total moles of electrons transferred at the anode must be equal to the total moles of electrons transferred at the cathode. Total electrons at anode = (electrons for H2S2O8H_2S_2O_8) + (electrons for O2O_2) Total electrons at cathode = (electrons for H2H_2)

So, 2×nH2S2O8+0.5 mol=1.0 mol2 \times n_{H_2S_2O_8} + 0.5 \text{ mol} = 1.0 \text{ mol} 2×nH2S2O8=1.0 mol−0.5 mol2 \times n_{H_2S_2O_8} = 1.0 \text{ mol} - 0.5 \text{ mol} 2×nH2S2O8=0.5 mol2 \times n_{H_2S_2O_8} = 0.5 \text{ mol} nH2S2O8=0.52=0.25 moln_{H_2S_2O_8} = \frac{0.5}{2} = 0.25 \text{ mol}

5. Calculate the weight of H2S2O8H_2S_2O_8 formed: First, calculate the molar mass of H2S2O8H_2S_2O_8: Molar mass = (2×Atomic mass of H)+(2×Atomic mass of S)+(8×Atomic mass of O)(2 \times \text{Atomic mass of H}) + (2 \times \text{Atomic mass of S}) + (8 \times \text{Atomic mass of O}) Molar mass = (2×1.008)+(2×32.06)+(8×16.00)(2 \times 1.008) + (2 \times 32.06) + (8 \times 16.00) Molar mass = 2.016+64.12+128.00=194.136 g/mol2.016 + 64.12 + 128.00 = 194.136 \text{ g/mol} (approximately 194 g/mol)

Weight of H2S2O8=nH2S2O8×Molar mass of H2S2O8H_2S_2O_8 = n_{H_2S_2O_8} \times \text{Molar mass of } H_2S_2O_8 Weight of H2S2O8=0.25 mol×194 g/mol=48.5 gH_2S_2O_8 = 0.25 \text{ mol} \times 194 \text{ g/mol} = 48.5 \text{ g}