Balance the redox reaction by inserting the appropriate coefficients. redox reaction: H^+ + CrO2^−4 + NO−2⟶Cr3+…
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Question “Balance the redox reaction by inserting the appropriate coefficients. redox reaction: H^+ + CrO2^−4 + NO−2⟶Cr3+…”
Balance the redox reaction by inserting the appropriate
coefficients.
redox reaction:
H^+ + CrO2^−4 + NO−2⟶Cr3+ + H2O + NO^−3
Answer
Cr in CrO4-2 is an oxidation state that has +6
The oxidation state for Cr in Cr+3 is +3.
So Cr in CrO4-2 becomes Cr+3.
N in NO2- has an oxidation state 3
N in NO3- is an oxidation state +5
Thus, NO2 in N is oxidized to NO3-
Half cell reduction:
CrO4-2 + 3e- –> Cr+3
Half cell for oxidation:
NO2- –> NO3- + 2e-
In both half reactions, balance number of electrons must be the same
Half cell reduction:
2 CrO4-2 + 6e- –> 2 Cr+3
Half cell for oxidation:
3 NO2- –> 3 NO3- + 6e-
Let’s combine them both.
2 CrO4-2 + 3 NO2- –> 2 Cr+3 + 3 NO3-
Add water to balance Oxygen
2 CrO4-2 + 3 NO2- –> 2 Cr+3 + 3 NO3- + 5 H2O
Add H+ to balance Hydrogen
2 CrO4-2 + 3 NO2- + 10 H+ –> 2 Cr+3 + 3 NO3- + 5 H2O
This is a balanced chemical equation in acidic media
Answer:
10 H^+ + 2 CrO2^-4 + 3 NO-2 2 Cr3+ + 5 H2O + 3 NO^-3
Conclusion
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