Question

Balance the following reaction that occurs in an acidic environment, UO2+ + Cr2O7^2- ----> UO2^2+ + Cr3+ I know that UO is reduced, and Cr is oxidized; so these are my half equations that are already balanced materially, I get confused when it comes to charge. UO2+ + H2O --->UO2^2+ + 2H Cr2O7^2- + 14H ---> 2Cr3+ +7H2O

Answer 1

Is it \[ UO_2 \ ^+ ->UO_2\ ^{2+} \] or \[ UO \ ^{2+} ->UO_2\ ^{2+} \] ?

Answer 2

second one :)

Answer 3

\[UO \ ^{2+} ->UO_2\ ^{2+}\]\[UO \ ^{2+} +H_2O->UO_2\ ^{2+}\]\[UO \ ^{2+} +H_2O->UO_2\ ^{2+}+2H^+\]\(UO \ ^{2+} +H_2O->UO_2\ ^{2+}+2H^++2e^-\) ---(1)

\[ Cr_2O_7\ ^{2-} -> 2Cr^{3+} \]\[ Cr_2O_7\ ^{2-} -> 2Cr^{3+} +7H_2O\]\[ Cr_2O_7\ ^{2-} +14H^+-> 2Cr^{3+} +7H_2O\]\( Cr_2O_7\ ^{2-} +14H^++6e^--> 2Cr^{3+} +7H_2O\) ---(2)

(1)x3 +(2)
\[3(UO \ ^{2+} +H_2O) + ( Cr_2O_7\ ^{2-} +14H^++6e^-)\]\[->3(UO_2\ ^{2+}+2H^++2e^-)+( 2Cr^{3+} +7H_2O)\]
\[3UO \ ^{2+} +3H_2O + Cr_2O_7\ ^{2-} +14H^+->3UO_2\ ^{2+}+6H^++ 2Cr^{3+} +7H_2O\]\[3UO \ ^{2+} + Cr_2O_7\ ^{2-} +8H^+->3UO_2\ ^{2+}+ 2Cr^{3+} +4H_2O\]

Answer 4

thank you!

Answer 5

Is that clear to you?!

Answer 6

yes, I though that I had to get both sides to zero charge but I see that that was not necessary. :)

Answer 7

Glad that you understand my workings!