Question

Pbo_2(8) + I^-(aq) rightarrow Pb^2+(aq) + I_2(s) I_2(s) rightarrow I^-(aq)
Balance the reaction in acidic conditions

Answer 1

PbO2(s) + 4H+(aq) + 2e- --> Pb+2(aq) + 2H2O(l)

2Cl-(aq) --> Cl2(g) + 2e-

add

PbO2(s) + 4H+(aq) + 2Cl-(aq) ------> Pb+2(aq) + Cl2(g) + 2H2O(l)

Make 4H+ -------------> 4HCl and add 2Cl- to other side becomes

PbO2(s) + 4HCl(aq) ---------> PbCl2(aq) + Cl2(g) + 2H2O(l)

Thank you and good luck.