Balance redox chemical reaction in acidic mediumCr2O72- + NO2- --> Cr3+ + NO3- (acid)I need full explanation about this

Expert Answers
sanjeetmanna eNotes educator| Certified Educator

`(Cr_2O_7^2^-) + (NO_2^-) ->(Cr^3^+) + (NO_3^-)` (acid)

First split the reaction into two half reaction with one half reaction represent the oxidation reaction and other half represent reduction reaction.

`(Cr_2O_7^2^-) ->(Cr^3^+)`

`(NO_2^-) -> (NO_3^-)`

Balance atoms other than oxygen.

`(Cr_2O_7^2^-) -> 2 (Cr^3^+)`

`(NO_2^-) -> (NO_3^-)`

 

In acidic medium add same number of H2O molecules to the side where there is deficient in oxygen to balance oxygen.

`(Cr_2O_7^2^-)->2(Cr^3^+) + 7(H_2O)`

`(H_2O) + (NO_2^-) ->(NO_3^-)`

 

Add H+ to the side where there is deficient in Hydrogen atom.

`14 (H^+) + (Cr_2O_7^2^-)->2 (Cr^3^+) + 7(H_2O)`

`(H_2O) + (NO_2^-) ->(NO_3^-) + 2 (H^+)`

 

Balance the charge by finding the net charge on both the sides and add as much as electron required to more positive side of the reaction.

`6(e^-) + 14 (H^+) + (Cr_2O_7^2^-)->2 (Cr^3^+) + 7 (H_2O)`

`(H_2O) + (NO_2^-) ->(NO_3^-) + 2 (H^+) + 2 (e^-)`

 

Both mass and charges are balanced now add up both the reactions.

Before adding the reactions number of electron gained must be equal to number of electrons lost. Multiply with the suitable number to get same number of electron on both the reaction.

`(6(e^-) + 14 (H^+) + (Cr_2O_7^2^-)->2 (Cr^3^+) + 7 (H_2O)) * 1`

`((H_2O) + (NO_2^-) ->(NO_3^-) + 2 (H^+) + 2 (e^-)) * 3`

 

`6(e^-) + 14 (H^+) + (Cr_2O_7^2^-) ->2 (Cr^3^+) + 7 (H_2O)`

`3 (H_2O) + 3 (NO_2^-) ->3 (NO_3^-) + 6 (H^+) + 6 (e^-)`

 

Add up both the reactions.

`6(e^-) + 14 (H^+) + (Cr_2O_7^2^-) ->2 (Cr^3^+) + 7 (H_2O)`

`3 (H_2O) + 3 (NO_2^-) ->3 (NO_3^-) + 6 (H^+) + 6 (e^-)`

---------------------------------------------------------

`8(H^+) + (Cr_2O_7^2^-) + 3(NO_2^-) -> 2(Cr^3^+) + 3(NO_3^-) + 4(H_2O)`

liajferguson eNotes educator| Certified Educator

When balancing a redox reaction, you should follow these steps.

1. Split the reaction into two half reactions.

2. Balance all atoms other than oxygen and hydrogen.

3. Balance the oxygen by adding H2O molecules.

4. Balance the hydrogen by adding H+ ions.

5. Balance the charges by adding electrons.

6. Make the number of electrons equal.

7. Add the half reactions together.

8. Simplify as much as possible.

For this problem Cr is being reduced, and N is being oxidized, so we can write these half reactions as step 1, shown in the image below. Balancing Cr and N, we get step 2. To balance oxygen, we can add H2O to get step 3. Then to balance hydrogen, add H+ to get step 4. Add electrons to balance the charges on both sides to get step 5. Balance the electrons gained with electrons lost by multiplying the reactions by, in this case, a factor of 3 to get step 6. Add the two half reactions back together to get step 7. And finally simplify to get your final answer in step 8.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
lochana2500 | Student

✪ Cr2O7²⁻ → Cr³⁺

6e⁻+ 14H⁺+Cr2O7²⁻ → 2Cr³⁺ + 7H2O ⇒ ❶

✪ NO2⁻ → NO3⁻

H2O + NO2⁻ → NO3⁻ + 2H⁺ + 2e⁻ ⇒ ❷

❶ + ❷x3 ⇒

8H⁺ + Cr2O7²⁻ + 3NO2⁻ → 2Cr³⁺ + 3NO3⁻ + 4H2O