balance these half reaction with permanganate and calculate the resulting oxidation number on the manganese ions:
a, MnO4^1-(aq) --->Mn2+(aq)
b,MnO4^1- (aq)--> Mn3+(aq)
c,MnO4^1-(aq)-->MnO2(S)
d,MnO4^1-(aq)-->MnO4^2-(aq)
a, MnO4-(aq) --->Mn2+(aq)
add H2O:
MnO4-(aq) --->Mn2+(aq) + 4H2O
add H+
8H+ + MnO4-(aq) --->Mn2+(aq) + 4H2O
balance charge
5e- + 8H+ + MnO4-(aq) --->Mn2+(aq) + 4H2O
the Mn goes from +7 to +2, gains 5 electrons
b)
MnO4- --> Mn+3
add H2O
MnO4- --> Mn+3 + 4H2O
add H+
8H + MnO4- --> Mn+3 + 4H2O
add electrons
4e- + 8H+ + MnO4- --> Mn+3 + 4H2O
Mn goes from +7 to +3, gains 4 electrons
c)
MnO4- --> MnO2
balanc eO
MnO4- --> MnO2 + 2H2O
balance H+
4H+ + MnO4- --> MnO2 + 2H2O
balance e-
3e- + 4H+ + MnO4- --> MnO2 + 2H2O
Mn goes form +5 to +4, it gains 1 electrons
d.
MnO4- --> MnO2-2
add H2O
MnO4- --> MnO2-2 + 2H2O
add H+
4H+ + MnO4- --> MnO2-2 + 2H2O
add electrons
4H+ + MnO4- --> MnO2-2 + 2H2O
5e- + 4H+ + MnO4- --> MnO2-2 + 2H2O
Mn goes from +7 to +2
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