Question

Determine the oxidation number of manganese in the products as per given equation. H+ + 2H2O + 2 MnO4^- + 5SO2 --> Products (in acidic solution)

Answer 1

Answer:

Explanation:

Permanganate ion is reduced from  

M

n

(

V

I

I

+

)

to  

M

n

(

I

V

+

)

in the oxide.

Explanation:

The reduction equation is:

M

n

O

−

4

+

4

H

+

+

3

e

−

→

M

n

O

2

+

2

H

2

O

This is balanced with respect to mass and charge. However, permanganate is normally reduced to  

M

n

2

+

, which is almost colourless. In fact, this reduction reaction is self indicating in that permanganate is intensely coloured.

Answer 2

Answer : The oxidation number of manganese in the product side is, (+2)

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The oxidation-reduction half reaction will be :

Oxidation : $SO_2+2H_2O\rightarrow SO_4^{2-}+4H^++2e^-$

Reduction : $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

In order of balance the electrons, we multiply the equation 1 by 5 and equation 2 by 2, we get:

The oxidation-reduction half reaction will be :

Oxidation : $5SO_2+10H_2O\rightarrow 5SO_4^{2-}+20H^++10e^-$

Reduction : $2MnO_4^-+16H^++10e^-\rightarrow 2Mn^{2+}+8H_2O$

Now adding both the equation, we get:

The overall reaction will be,

$2MnO_4^-+5SO_2+2H_2O\rightarrow 2Mn^{2+}+5SO_4^{2-}+4H^+$

Hence, the oxidation number of manganese in the product side is, (+2)