Question

HNO3+S->H2SO4+NO.

Determine how many electrons are involved in each half-reaction. How many electrons should be assigned to which side of this half-reaction equation?

Answer 1

In the oxidation half-reaction, 3 electrons are involved, while in the reduction half-reaction, 6 electrons are involved. The electrons are assigned to each side of the equation to balance them.

Step-by-step explanation:

The half-reaction equation given is HNO3 + S → H2SO4 + NO. In this equation, the oxidation state of nitrogen (N) changes from +5 in HNO3 to +2 in NO. Therefore, 3 electrons are involved in the oxidation half-reaction (HNO3 → NO). On the other hand, the oxidation state of sulfur (S) changes from 0 in S to +6 in H2SO4. Hence, 6 electrons are involved in the reduction half-reaction (S → H2SO4).

In order to balance the electrons on both sides of the equation, we assign 6 electrons to the left side of the equation (oxidation half-reaction) and 3 electrons to the right side of the equation (reduction half-reaction).