Redox reactions

Thanx.

Plz anyone solve the 2nd one

Cr₂O₇ +NO₂^- - -> Cr³⁺ +NO₃^- (Neutral medium)
Let us take first part:
Cr₂O₇ →Cr³⁺
Cr₂O₇ →2Cr³⁺ (Balancing chromium on both sides)
Cr₂O₇ →2Cr³⁺+7H₂O (7 water molecule added to oxygen deficient side)
Cr₂O₇ +14H⁺ →2Cr³⁺+7H₂O (to balance hydrogen-14 H⁺added to hydrogen deficient side)
Cr₂O₇ +14H⁺+8e^- →2Cr³⁺+7H₂O .....(1) (to balance charges 14⁺ & 6⁺, 8 e^- is added to LHS)

Now let's take the 2nd part:
NO₂^- →NO₃^-
NO₂^-+H₂O →NO₃^- (1 water molecule added to oxygen deficient side )
NO₂^-+H₂O →NO₃^- +2H⁺ (to balance hydrogen-2 H⁺added to hydrogen deficient side)
NO₂^-+H₂O -2e^- →NO₃^- +2H⁺ ......(2) (to balance charges 1^- & 1⁺, 2 e^- is subtracted from LHS)

To equalize the number of electrons gained by (1)st part, the (2)nd part must be multiplied by 4, so (2) becomes

4NO₂^-+4H₂O -8e^- →4NO₃^- +8H⁺

Now a solution of simultaneous linear equation between (1) & (2) :-

Cr₂O₇ +14H⁺+8e^- →2Cr³⁺+7H₂O
4NO₂^-+4H₂O -8e^- →4NO₃^- +8H⁺

On adding (1) & (2) :-

On cancelling the electrons-

Cr₂O₇ +14H⁺ →2Cr³⁺+7H₂O
4NO₂^-+4H₂O →4NO₃^- +8H⁺

On cancelling H₂O molecules-

Cr₂O₇ +14H⁺ →2Cr³⁺+3H₂O
4NO₂^- →4NO₃^- +8H⁺

On cancelling the H⁺ ions-

Cr₂O₇ +6H⁺ →2Cr³⁺+3H₂O
4NO₂^- →4NO₃^-

Final addition of reactants & products:-

Cr₂O₇ +6H⁺ + 4NO₂^-→2Cr³⁺+3H₂O + 4NO₃^- (The equation is balanced and charge on both sides are equal that is 2⁺)