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   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

a) the particles of the chair are strongly held together.