Find reducing and oxidizing agents

SO4^2- is the oxidizing agent.

l^- is the reducing agent.

Explanation:

SO4^2- + I^- + H^+ —> H2S + I2 + H2O

To know which is the oxiding agent or the reducing agent, let us calculate the change in oxidation number of each element. This is illustrated below:

Note: the oxidation number of Hydrogen (H) is always +1 except in hydride where it is -1.

The oxidation number of Oxygen (O) is always -2 except in peroxide where it -1

For S:

SO4 = -2

S + 4O = -2

O = - 2

S =.?

S + (4 x -2) = -2

S - 8 = -2

Collect like terms

S = -2 + 8

S = +6

H2S = 0

2H + S = 0

H = +1

S =..?

2(1) + S = 0

2 + S = 0

Collect like terms

S = 0 - 2

S = -2

The oxidation number of S changes from +6 to -2

For I:

I = - 1

I = 0

The oxidation of I changes from -1 to 0.

Since the oxidation number of S changes from +6 to -2 i.e reduce, therefore SO4^2- is the oxidizing agent.

The oxidation number of I changes from -1 to 0 ie increased. Therefore, l^- is the reducing agent.

the answer is minerals.