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.

alienation, transformation, consumption, and the relativity of truth.

you can choose any one of these, they are all important themes.

these themes reflect the distinct sensibilities of both the modernist and postmodern aesthetic movements. (from google, you should paraphrase)

hope this !

- melanie