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

sanjeetmanna | College Teacher | (Level 3) Assistant Educator

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`(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^-)`

`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^-)`

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

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

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✪ Cr2O7²⁻ → Cr³⁺

6e⁻+ 14H⁺+Cr2O7²⁻ → 2Cr³⁺ + 7H2O ⇒ ❶

✪ NO2⁻ → NO3⁻

H2O + NO2⁻ → NO3⁻ + 2H⁺ + 2e⁻ ⇒ ❷

❶ + ❷x3 ⇒

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