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Assign oxidation states to each of the atoms in the reactants and products listed below. Then...

Assign oxidation states to each of the atoms in the reactants and products listed below. Then balance the equation using the oxidation state method. What is being oxidized and reduced? What is the oxidizing agent and the reducing agent?

H₂C₂O₄(aq) + MnO₄^1-(aq) + H⁺(aq) ---> CO₂(aq) + Mn²⁺(aq) + H₂O(l)

Expert Solution

unbalanced equation

H2C2O4(aq) + MnO4-(aq) + H+(aq)? CO2(aq) + Mn2+(aq) + H2O(l)

Write oxidation number of each atom

half reactions are

O represents oxidation

R represents reduction

Mn is reduced, Mn is oxidizing agent

C is oxidized, C is reducing agent

Balance the atoms except H and O

O:

H2C2O4 ? 2CO2

R:

MnO4- ? Mn2+

Balance the oxygen atoms

O:

H2C2O4 ? 2CO2

R:

MnO4- ? Mn2+ + 4H2O

Balance the hydrogen atoms

by adding protons (H+).

O:

H2C2O4 ? 2CO2 + 2H+

R:

MnO4- + 8H+ ? Mn2+ + 4H2O

Balance the charge

O:

H2C2O4 ? 2CO2 + 2H+ + 2e-

R:

MnO4- + 8H+ + 5e- ? Mn2+ + 4H2O

Electron Balance

O:

H2C2O4 ? 2CO2 + 2H+ + 2e-

Multiply by 5

R:

MnO4- + 8H+ + 5e- ? Mn2+ + 4H2O

Multiply by 2

O:

5H2C2O4 ? 10CO2 + 10H+ + 10e-

R:

2MnO4- + 16H+ + 10e- ? 2Mn2+ + 8H2O

Add the half-reactions together.

5H2C2O4 + 2MnO4- + 16H+ + 10e- ?10CO2 + 2Mn2+ + 10H+ + 8H2O + 10e-

Simplify the equation.

5H2C2O4 + 2MnO4- + 6H+ ? 10CO2 + 2Mn2+ + 8H2O

Amanda K answered 2 years ago

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