Complete and balance the following redox reaction in acidic solution

Question

TeO3^-2 (aq) +N2O4(g)-> Te(s) + NO3^- (aq)

Answer 1

TeO₃^2-(aq) + N₂O₄(g) ==> Te(s) + NO₃^-(aq) .. unbalanced

Split into 2 half reactions:

TeO₃^2-(aq) ==> Te(s) .. reduction half reaction (Te goes from +4 O.N. to 0 O.N.)

N₂O₄(g) ==> NO₃^-(aq) .. oxidation half reaction (N goes from +4 O.N. to +5 O.N.)

Balance each in acidic solution:

TeO₃^2-(aq) ==> Te(s) + 3H₂O(l) .. balanced for Te and O

TeO₃^2-(aq) + 6H⁺(aq) ==> Te(s) + 3H₂O(l) .. balanced for Te, O and H (using acid, H⁺)

TeO₃^2-(aq) + 6H⁺(aq) + 4e- ==> Te(s) + 3H₂O(l) .. balanced for Te, O, H and Charge = BALANCED RXN

N₂O₄(g) ==> 2NO₃^-(aq) .. balanced for N

N₂O₄(g) + 2H₂O(l) ==> 2NO₃^-(aq) .. balanced for N and O

N₂O₄(g) + 2H₂O(l) ==> 2NO₃^-(aq) + 4H⁺(aq) .. balanced for N, O and H (using acid, H⁺)

N₂O₄(g) + 2H₂O(l) ==> 2NO₃^-(aq) + 4H⁺(aq) + 2e- .. balanced for N, O, H and charge = BALANCED RXN

Multiply oxidation oxidation rxn by 2 to equalize electrons, add them and then combine/cancel like terms:

TeO₃^2-(aq) + 6H⁺(aq) + 4e- ==> Te(s) + 3H₂O(l)

2N₂O₄(g) + 4H₂O(l) ==> 4NO₃^-(aq) + 8H⁺(aq) + 4e-

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TeO₃^2-(aq) + 2N₂O₄(g) + H₂O(l) ==> Te(s) + 4NO₃^-(aq) + 2H⁺(aq) .. BALANCED REDOX