Question

Balance the following redox equation in basic conditions:

Ag(s) + MnO₄⁻(aq) ⟶ MnO₂(s) + Ag⁺(aq)

What is the stochiometric coefficient for H₂O in the balanced equation?

Answer 1

Ag(s) + MnO4⁻(aq) ⟶ MnO2(s) + Ag+(aq)

split it int oxidation and reduction half

OX: Ag --> Ag+
RED: MnO4- --> MnO2

balance O atoms by adding H2O

OX: Ag --> Ag+
RED: MnO4- --> MnO2 +2H2O

to balance H atom by adding protons

OX: Ag --> Ag+
RED: MnO4- + 4H+ --> MnO2 +2H2O

since the reaction is basic medium add OH to cancel
H+ on both side

OX: Ag --> Ag+
RED: MnO4- + 4H2O --> MnO2 +2H2O +2OH-

balance charge by adding electrons

OX: Ag --> Ag+ e-
RED: MnO4- + 4H2O +3e- --> MnO2 +2H2O +2OH-

make electron lost = electron gained , so multiply OX;
half by 3

OX:3 Ag -->3 Ag+ 3e-
RED: MnO4- + 4H2O +3e- --> MnO2 +2H2O +2OH-

Add both half reactions cancel same species on opposite
side of the arrow we get overall reaction

3Ag(s) + MnO4-(aq) + 4H2O(l) + 3e- --> 3Ag+(aq) + MnO2(s) + 3e- + 2H2O(l) + 4OH-(aq)

answer
3Ag(s) + MnO4-(aq) + 2H2O(l) → 3Ag+(aq) + MnO2(s) + 4OH-(aq)