2 MnO4−(aq) + 10 Cl−(aq) + 16 H+(aq) → 5 Cl2(g) + 2 Mn2+(aq) + 8 H2O(l) Oxidizing agent: Reducing agent:

MnO4 is the oxidizing agent.

Cl is the reducing agent.

Explanation:

To know which is oxidizing agent and which is reducing agent, let us calculate the change in oxidation number of each element in the equation below:

2MnO4−(aq) + 10Cl−(aq) + 16H+(aq) → 5Cl2(g) + 2Mn2+(aq) + 8H2O(l)

Please note:

1. The oxidation state of oxygen is always –2 except in peroxide where it is –1

2. The oxidation state of Hydrogen is always +1 except in hydrides where it is –1

3. The oxidation state of chlorine is always –1

For Mn:

At left hand side

MnO4 = –1

Mn + 4O = –1

O = –2

Mn + (4 x –2) = –1

Mn – 8 = –1

Collect like terms

Mn = –1 + 8

Mn = +7

At the right hand side, Mn is +2.

The oxidation state of Mn changes from +7 to +2.

For Cl:

At the left hand side

Cl = –1

At the right hand side

Cl = 0

The oxidation state of Cl changes from – 1 to 0.

The oxidation state of O and H are unchanged.

Since the oxidation state of Mn changes from +7 to +2 i.e reduced, MnO4 is the oxidizing agent.

Since the oxidation state of Cl changes from – 1 to 0 i.e increase, Cl is the reducing agent.

the answer is d.) potassium