#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.
