Question #a2c32

1 Answer
May 14, 2017

Answer:

#Cr_2O_7^(2-) (aq)# + #3SO_2 (g)# + #2H^+ (aq)##2Cr^(3+)(aq)# + #3SO_4^(2-) (aq)# + #H_2O (l)#

Explanation:

#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.