2HNO 3(aq) + SO2(g) → H2SO4(aq) + 2NO2(g)
2KMnO 4(aq) + 5SO2(g) + 2H2O → K2SO4(aq) +MnSO4(aq) 2H2SO4(aq)
FeCI3(aq) + SO2(g) + 2H2O(l) → FeCI 2(g)+ 2HCI(g) + H2SO4(aq)
2H2S(g) + SO2(g) →2H2O(l) + 3S(s)
Explanation
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Discussions (6)
A.
Sulfur goes from +4 (in SO_2) to +6 (in H_2SO_4).
Sulfur is oxidized; therefore, SO_2 is acting as a reducing agent.
B.
Sulfur goes from +4 to +6 (in the sulfate ions).
Sulfur is oxidized; SO_2 is a reducing agent.
C.
Sulfur goes from +4 to +6.
Sulfur is oxidized; SO_2 is a reducing agent.
D.
In SO_2, sulfur is +4.
In the product (S), sulfur is 0.
The oxidation state decreased (+4 to 0). This is reduction.
Conclusion
In reaction D, SO_2 removes electrons from H_2S (where sulfur is -2) to turn it into elemental sulfur, while the sulfur in SO_2 itself is reduced.
Correct Answer: D
