Question

mn+bio3=mno4+bi como balancear en medio ácido metodo ion electron

Answer 1

                ₊₅ ₋₂                       ₊₇   ₋₂

   Mn⁺² + (BiO3)⁻¹  + (H)⁺¹  → (MnO4)⁻¹ + (Bi)⁺³

                             (Mn)⁺² → (MnO4)⁻¹

                          (BiO3)⁻¹ →    (Bi)⁺³

                        ````````````````````````````````````````````

                  (Mn)⁺² + 4H2O → (MnO4)⁻¹  + 8(H⁺) + 5⁻e            Ι 2

      2e⁻ +  6(H⁺) + (BiO3)⁻¹ →    (Bi)⁺³ +3H2O                          Ι 5

    ````````````````````````````````````````````````````````````````````````````````````````````````````````

           2(Mn)⁺² + 8H2O → 2(MnO4)⁻¹  + 16(H⁺) +  10e⁻           sumar las

 10e⁻ +  30(H⁺) + 5(BiO3)⁻¹ →   5(Bi)⁺³ +  15H2O                      semirreacciones

``````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````

2(Mn)⁺²+8H2O +10e⁻+30(H⁺)+5(BiO3)⁻¹→2(MnO4)⁻¹+16(H⁺) + 10e⁻+5(Bi)⁺³+ 15H2O

igualar las semirreacciones

                 2(Mn)⁺² + 5(BiO3)⁻¹ + 14(H⁺)  →  5(Bi)⁺³+ 7H2O (balanceada)