SO_2 gas is bubbled through a solution of K_2Cr_2O_7
Half equations:
Cr_2O_7^(2-) (aq) + 14H^+ (aq) + 6e^- →2Cr^(3+) (aq) + 7H_2O (l)
SO_2 (g) + 2H_2O (l) → SO_4^(2-) (aq) + 4H^+ (aq) + 2e^-
(i'm not sure if you need this but i'll just include it)
Multiply half-eq'n 2 by 3
3SO_2 (g) + 6H_2O (l) → 3SO_4^(2-) (aq) + 12H^+ (aq) + 6e^-
Remove electrons from both equations and then add them together:
Cr_2O_7^(2-) + 14H^+ + 3SO_2 + 6H_2O →2Cr^(3+) + 7H_2O + 3SO_4^(2-) + 12H^+
And basically just subtract the same species to get a simpler answer.
e.g. 14H^+ on RHS minus 12H^+ on LHS gives 2H^+ on RHS.
Full equation:
Cr_2O_7^(2-) (aq) + 3SO_2 (g) + 2H^+ (aq) → 2Cr^(3+)(aq) + 3SO_4^(2-) (aq) + H_2O (l)
Solution turns from orange to green.
