# Balancing a redox basic solution?

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Balance the following redox reaction in basic (alkaline) solution and provide the correct coefficients for the indicated species as your answers.

H2O2(aq) + Mn2+(aq) → MnO2(s)+ H2O(l)

You do NOT need to enter the coefficients for H2O, H+ or OH–.

• 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