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Question: Perdisulphuric acid, H₂S₂O₈ can be prepared by electrolytic oxidation of H₂SO₄, oxygen and hydrogen ...

Perdisulphuric acid, H₂S₂O₈ can be prepared by electrolytic oxidation of H₂SO₄, oxygen and hydrogen gases are by products. In such an electrolysis, 9.08 L of H₂ and 2.27 L of O₂ were generated at STP. What is the mass of H₂S₂O₈ formed?

A

0

B

77.6 g

C

38.8 g

D

19.4 g

Answer

38.8 g

Explanation

Solution

The electrolysis process involves the following electrode reactions: Anode:

  1. Formation of perdisulphuric acid: 2H2SO4→H2S2O8+2H++2e−2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-
  2. Formation of oxygen gas: 2H2O→O2+4H++4e−2H_2O \rightarrow O_2 + 4H^+ + 4e^- Cathode:
  3. Formation of hydrogen gas: 2H++2e−→H22H^+ + 2e^- \rightarrow H_2

Let nH2S2O8n_{H_2S_2O_8}, nO2n_{O_2}, and nH2n_{H_2} be the moles of H₂S₂O₈, O₂, and H₂ produced, respectively. The number of moles of electrons transferred for each process are:

  • For H₂S₂O₈: 2×nH2S2O82 \times n_{H_2S_2O_8}
  • For O₂: 4×nO24 \times n_{O_2}
  • For H₂: 2×nH22 \times n_{H_2}

The total electrons produced at the anode must equal the total electrons consumed at the cathode. Electrons from anode = Electrons for H₂S₂O₈ formation + Electrons for O₂ formation Electrons for cathode = Electrons for H₂ formation

Thus, 2×nH2S2O8+4×nO2=2×nH22 \times n_{H_2S_2O_8} + 4 \times n_{O_2} = 2 \times n_{H_2}. Dividing by 2, we get: nH2S2O8+2×nO2=nH2n_{H_2S_2O_8} + 2 \times n_{O_2} = n_{H_2}. Rearranging to find moles of H₂S₂O₈: nH2S2O8=nH2−2×nO2n_{H_2S_2O_8} = n_{H_2} - 2 \times n_{O_2}.

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 L. Given volumes: VH2=9.08V_{H_2} = 9.08 L and VO2=2.27V_{O_2} = 2.27 L.

Moles of H₂: nH2=VH222.4=9.0822.4n_{H_2} = \frac{V_{H_2}}{22.4} = \frac{9.08}{22.4} mol. Moles of O₂: nO2=VO222.4=2.2722.4n_{O_2} = \frac{V_{O_2}}{22.4} = \frac{2.27}{22.4} mol.

Substitute these values into the equation for nH2S2O8n_{H_2S_2O_8}: nH2S2O8=9.0822.4−2×2.2722.4n_{H_2S_2O_8} = \frac{9.08}{22.4} - 2 \times \frac{2.27}{22.4} nH2S2O8=9.08−4.5422.4n_{H_2S_2O_8} = \frac{9.08 - 4.54}{22.4} nH2S2O8=4.5422.4n_{H_2S_2O_8} = \frac{4.54}{22.4} mol.

The molar mass of H₂S₂O₈ is approximately 194 g/mol. Mass of H₂S₂O₈ = nH2S2O8×Molar Massn_{H_2S_2O_8} \times \text{Molar Mass} Mass = 4.5422.4×194≈39.32\frac{4.54}{22.4} \times 194 \approx 39.32 g.

If NTP conditions (molar volume 22.7 L/mol) are used instead of STP (22.4 L/mol): nH2S2O8=4.5422.7n_{H_2S_2O_8} = \frac{4.54}{22.7} mol. Mass = 4.5422.7×194≈0.200×194=38.8\frac{4.54}{22.7} \times 194 \approx 0.200 \times 194 = 38.8 g. This matches option (c).