Balance in acidic soln:Tin(II) is oxidized to tin(IV) with hydrogen peroxide, Sn2+(aq)+H2O2(aq)=H2O(l)+Sn4+?c

Tin (II) is oxidized to tin (IV) with hydrogen peroxide: Sn2+ (aq) + H2O2 (aq) -> H2O (l) + Sn 4+ (aq). Balance the equation in an acidic solution.


Sn2+ -----------> Sn4+ + 2e- oxidation

H2O2 + 2H+ + 2e- ---------> 2H2O reduction 

adding both equation, gives
Sn2+ + H2O2 + 2H+ ------------> 2 H2O + Sn4+