Balancing a redox basic solution?

2 Answers

• What you do is balance in acid, then add hydroxides.

  Here’s one half-reaction:

  Mn2+ —> MnO2

  2H2O + Mn2+ —> MnO2

  2H2O + Mn2+ —> MnO2 + 4H+

  2H2O + Mn2+ —> MnO2 + 4H+ + 2e-

  add 4 hydroxides to each side:

  4OH- + 2H2O + Mn2+ —> MnO2 + 4H2O + 2e-

  On the right side, I combined the H+ and the OH-

  Eliminate water:

  4OH- + Mn2+ —> MnO2 + 2H2O + 2e-

  Here’s the other half-reaction:

  H2O2 —> H2O

  H2O2 —> 2H2O

  2H+ + H2O2 —> 2H2O

  2e- + 2H+ + H2O2 —> 2H2O

  2e- + 2H2O + H2O2 —> 2H2O + 2OH-

  2e- + H2O2 —> 2OH-

  You may do the rest. Email me if stuck. Here’s the redox portion of my web site:

  http://www.chemteam.info/Redox/Redox.html

• Mn2(aq) ——–> MnO2(s) Oxidation

  H2O2(aq) ——> H2O(l) Reduction

  Step 1: Balance all elements except for H and O

  Mn2(aq) ——–> 2MnO2(s)

  H2O2(aq) ——> H2O(l)

  Step 2: Balance O by adding H2O

  Mn2(aq) + 2H2O ——–> 2MnO2(s)

  H2O2(aq) ——> H2O(l) + H2O

  Step 3: Balance H by adding H+

  Mn2(aq) + 2H2O ——–> 2MnO2(s) + 4H

  H2O2(aq) + 2H+ ——> H2O(l) + H2O

  Step 4: Neutralize H+ by adding enough OH- to neutralize each H+. Add the same number of OH- ions to each side of the equation.

  Mn2(aq) + 2H2O+ 4HO- ——–> 2MnO2(s) + 4H + 4HO-

  H2O2(aq) + 2H + 2HO- ——> H2O(l) + H2O + 2OH

  which becomes

  Mn2(aq) + 2H2O+ 4HO- ——–> 2MnO2(s) + 4H2O

  H2O2(aq) + 2H2O ——> 2H2O(l) + 2OH

  Step 5: Balance charges

  Mn2(aq) + 2H2O+ 4HO ——–> 2MnO2(s) + 4H2O + 2e-

  H2O2(aq) + 2H2O + 2e- ——> 2H2O(l) + 2OH

  Step 6: Add the half-reaction together,

  Mn2(aq) + 2H2O+ 4HO + H2O2(aq) + 2H2O + 2e- —> 2MnO2(s) + 4H2O + 2e- + 2H2O(l) + 2OH

  and combine like terms, leaving you with your answer.

  Mn2(aq) + H2O2(aq) + 2OH(aq) ————–> 2MnO2(s) + 2H2O(l)

  Bill