Balance the equation Run in an acidic solution: MnO2+Cu+2 + H+-->MnO4-+Cu++H2O

Question

Balance the equation Run in an acidic solution:

(Not OH-)

MnO2+Cu+2 + H+-->MnO4-+Cu++H2O

Answer 1

Use the half reaction method:

MnO₂ ==> MnO₄^- ... oxidation half reaction

MnO₂ + 2H₂O ==> MnO₄^- ... balanced for Mn and O

MnO₂ + 2H₂O ==> MnO₄^- + 4H⁺ ... balanced for Mn, O and H using acid (H⁺)

MnO₂ + 2H₂O ==> MnO₄^- + 4H⁺ + 3e- balanced for Mn, O and H and charge = BALANCED OX. RXN

Cu²⁺ ==> Cu ... reduction half reaction

Cu²⁺ + 2e- ==> Cu ... balanced for Cu and charge = BALANCED RED.RXN

To equalize the electrons we must multiply the oxidation rxn by 2 and the reduction rxn by 3. Then add the two equations together and combine/cancel like terms.

2MnO₂ + 4H₂O ==> 2MnO₄^- + 8H⁺ + 6e-

3Cu²⁺ + 6e- ==> 3Cu

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

2MnO₂ + 4H₂O + 3Cu²⁺ ==> 2MnO₄^- + 8H⁺ + 3Cu BALANCED REDOX EQUATION