In a neutral medium, MnO4- ions reduces into MnO2 while oxidizing C2O4 2- into CO2. What is the stoichiometric ratio MnO4-:C2O4 2- in the balanced chemical reaction? A) 2:5 B) 5:2 C) 3:2 D) 2:3 E) None of the above answers are correct.
Question
In a neutral medium, MnO4- ions reduces into MnO2 while oxidizing C2O4 2- into CO2. What is the stoichiometric ratio MnO4-:C2O4 2- in the balanced chemical reaction?
A) 2:5 B) 5:2 C) 3:2 D) 2:3 E) None of the above answers are correct.
Solution
To find the stoichiometric ratio between MnO4- and C2O4 2-, we need to balance the redox reaction.
The half-reaction for the reduction of MnO4- to MnO2 is:
MnO4- + 2H2O → MnO2 + 4OH-
This shows that each MnO4- ion gains 3 electrons (e-) to become MnO2.
The half-reaction for the oxidation of C2O4 2- to CO2 is:
C2O4 2- → 2CO2 + 2e-
This shows that each C2O4 2- ion loses 2 electrons to become CO2.
To balance the electrons gained and lost in the two half-reactions, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
3[C2O4 2- → 2CO2 + 2e-] 2[MnO4- + 2H2O → MnO2 + 4OH-]
This gives us the balanced redox reaction:
3C2O4 2- + 2MnO4- + 4H2O → 6CO2 + 2MnO2 + 8OH-
So, the stoichiometric ratio of MnO4- to C2O4 2- is 2:3.
Therefore, the correct answer is D) 2:3.
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