Determine the oxidation number of manganese in the products as per given equation. H+ + 2H2O + 2 MnO4^- + 5SO2 --> Products (in acidic solution)
Answers
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 : 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 :
Reduction :
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 :
Reduction :
Now adding both the equation, we get:
The overall reaction will be,
Hence, the oxidation number of manganese in the product side is, (+2)