For a particular redox reaction SO32– is oxidized to SO42– and Ag is reduced to Ag. Complete and balance the equation for this reaction in basic solution. Phases are optional.
(SO3)^2- + Ag+ –> (SO4)^-2 + Ag
Could you please provide detailed steps on how to balence this reaction in BASIC solution (the answer uses H2O and OH)
Expert Answer
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