Question: Balance the following oxidation-reduction reaction in basic condition:
H2O +(S2-)+(MnO4-)--> S+MnS+(OH-)
My answer is:
6H2O +5(S2-)+2(MnO4-)--> 5S+2MnS+12(OH-)
However, thats not an option.
The equation is not balanced , amount of Oxygen and sulfur is not correct.
Reduction
MnO4- + 4 H2O + 5 e- => Mn2+ + 8 OH-
Oxidation
S2- => S + 2 e-
Multiply each equation so the electrons equal
2 MnO4- + 8 H2O + 10 e- => 2 Mn2+ + 16 OH-
5 S2- => 5 S + 10 e-
Addition
2 MnO4- + 8 H2O + 5 S2- => 2 Mn2+ + 16 OH- + 5 S
Precipitation of MnS need two more S2-
2 MnO4- + 8 H2O + 7 S2- => 2 MnS + 16 OH- + 5 S
Great explanation !