Answer:
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.
