Balance the following redox reactions by ion-electron method (a) Mn04-(aq) + I-(aq) → MnO2 (s) + 12 (s) (in basic medium) (b) MnO4-(aq) + SO2 ( g) → Mn2 + (aq) + HSO4- (aq) (in acidic solution)
Answers
◇ ANSWER ◇
The unbalanced chemical equation is:
▪MnO4−(aq)+SO2(g)→Mn2+(aq)+HSO4−(aq)
The oxidation half reaction is:-
SO2(g)+2H2O(l)→HSO4−(aq)+3H+(aq)+2e−(aq).
The reduction half reaction is
MnO4−(aq)→Mn(2+)(aq).
In the reduction half reaction, the oxidation number of Mn changes from +7 to +2. Hence, 5 electrons are added to LHS of the reaction.
MnO4−(aq)+5e−→Mn2+(aq)
Charge is balanced in the reduction half reaction by adding 8 hydrogen ions to LHS.
MnO4−(aq)+5e−+8H+(aq)→Mn2+(aq)
To balance O atoms, 4 water molecules are added on RHS.
MnO4−(aq)+5e−+8H+(aq)→Mn2+(aq)+4H2O(l)
To equalize the number of electrons, the oxidation half reaction is multiplied by 5 and the reduction half reaction is multiplied by 2.
5SO2(g)+10H2O(l)→5HSO
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Answer:
In geometry, a hexagon can be defined as a polygon with six sides. The two-dimensional shape has 6 sides, 6 vertices and 6 angles.