Answer:
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.
