[SOLUTION] For a particular redox reaction SO32– is oxidized to SO42– and Ag is reduced to Ag. Complete…..

In basic solution = OH- must be present

split equations

SO3-2 = SO4-2

Ag+ = Ag

balanc O adding H2O

H2O + SO3-2 = SO4-2

Ag+ = Ag

balance H adding H+

H2O + SO3-2 = SO4-2 + 2H+

Ag+ = Ag

balance charge, both sides must have the same value

H2O + SO3-2 = SO4-2 + 2H+ + 2e-

e- + Ag+ = Ag

balance e-

H2O + SO3-2 = SO4-2 + 2H+ + 2e-

2e- + 2Ag+ = 2Ag

add all

2e- + 2Ag+ + H2O + SO3-2 = SO4-2 + 2H+ + 2e- +  2Ag

now cancel common terms

2Ag+ + H2O + SO3-2 = SO4-2 + 2H+ +2Ag

now, add OH- for basic conditions

2OH- + 2Ag+ + H2O + SO3-2 = SO4-2 + 2H+ +2Ag + 2OH-

add water

2OH- + 2Ag+ + H2O + SO3-2 = SO4-2 + +2Ag + 2H2O

cancel common terms

2OH- + 2Ag+ + SO3-2 = SO4-2 + +2Ag + H2O

add phases

2OH-(aq) + 2Ag+(aq) + SO3-2(aq) = SO4-2(aq) + +2Ag(s) + H2O(l)

this is now balanced in basic conditions

balance the following equation in basic conditions. phases are optional. so3