Question

Consider the following redox reaction:

5 H3AsO3 + H3O+ + 2 MnO4- →5 H2AsO4- + 2 Mn2+ + 4 H2O

Which species gets oxidized? .

. Which species gets reduced?

Which species is the oxidizing agent?

Which species is the reducing agent?

Which species gains electrons?

Which species loses electrons?

Answer 1

Oxidation number of each element in H3AsO3 is

H=+1

O=-2

As=+3

Oxidation number of each element in H3O+1 is

H=+1

O=-2

Oxidation number of each element in MnO4-1 is

O=-2

Mn=+7

Oxidation number of each element in H2AsO4-1 is

H=+1

O=-2

As=+5

Oxidation number of each element in Mn2+1 is

Mn=0

Oxidation number of each element in H2O is

H=+1

O=-2

In summary,

Oxidation number of each element in reactant is

H=+1

O=-2

As=+3

Mn=+7

Oxidation number of each element in product is

H=+1

O=-2

As=+5

Mn=0

1)

As in H3AsO3 has oxidation state of +3

As in H2AsO4- has oxidation state of +5

So, As in H3AsO3 is oxidised to H2AsO4-

Answer: As

2)

Mn in MnO4- has oxidation state of +7

Mn in Mn2+ has oxidation state of 0

So, Mn in MnO4- is reduced to Mn2+

Answer: Mn

3)

Since Mn in MnO4- is reduced

MnO4- is oxidising agent

Answer: MnO4-

4)

Since As in H3AsO3 is oxidised

H3AsO3 is reducing agent

Answer: H3AsO3

5)

Since Mn is reduced, it gains electrons

Answer: Mn

6)

Since As is oxidised, it loses electrons

Answer: As