Question

Write the balanced equation for the reaction between MnO4^- and C2)4^2-. Identify the oxidizing reagent and the reducing reagent. Identify the elements that undergo oxidation and reaction including their oxidation number changes. 2MnO4^- + 5 C2O4^2- + 16H^+ --> 2Mn^2+ + 10CO2 + 8H20

Answer 1

First, define the “ACIDIC” solution/conditions as H+ presence and

Basic solution implies OH- once it is balanced.

Also; note that ALL species must be balanced, as well as charges

Typical steps:

1) split half redox cells

MnO4- = Mn+2

C2O4-2 = CO2

2) balance atoms other than O,H

MnO4- = Mn+2

C2O4-2 = 2CO2

3) balance O by adding H2O

MnO4- = Mn+2 + 4H2O

C2O4-2 = 2CO2

4) balance H by adding H+

8H+ + MnO4- = Mn+2 + 4H2O

C2O4-2 = 2CO2

5) balance charge by adding e-

5e- + 8H+ + MnO4- = Mn+2 + 4H2O

C2O4-2 = 2CO2 + 2e-

6) balance e- by multiplying by the Greatest common divisor

10e- + 16H+ + 2MnO4- = 2Mn+2 + 8H2O

5C2O4-2 = 10CO2 + 10e-

7) Add both equations

5C2O4-2 + 10e- + 16H+ + 2MnO4- = 2Mn+2 + 8H2O + 10CO2 + 10e-

8) simplify repeating elements, H+, H2O, and e- typically

5C2O4-2 + 16H+ + 2MnO4- = 2Mn+2 + 8H2O + 10CO2

now..

Reduction = species that GAINS electrons

Oxidation = process in which a specie will LOSS electrons

Reducing agent = The species that favors reduction, i.e. it will oxidize in order to reduce another species

Oxidizing agent = The species that favors oxidation, i.e. it will reduce in order to oxidise another species

in this case

Mn goes from +7 to +2, then it gains electrons, therefore, it reduces, it is the oxidizing agent

C goes from +3 to +4, therefore, loses electrons, it gets oxidized, which implies it is the reducing agent