Basic concept doubts in redox reactions

Click For Summary
SUMMARY

The discussion centers on determining whether the reaction of oxalate ion ([C2O4]²⁻) to carbon dioxide (CO2) involves oxidation or reduction, and calculating the number of electrons involved. The oxidation states of carbon change from +3 on the left-hand side (LHS) to +4 on the right-hand side (RHS), indicating that carbon loses electrons. The correct interpretation reveals that 2 electrons are lost in the reaction, confirming that this process is oxidation. The balanced half-reaction is C2O42- → 2CO2 + 2e-.

PREREQUISITES
  • Understanding of oxidation states and how to calculate them
  • Familiarity with redox reactions and their components
  • Knowledge of balancing chemical equations
  • Experience with half-reaction methods in redox chemistry
NEXT STEPS
  • Study the principles of oxidation and reduction in redox reactions
  • Learn how to balance redox reactions using half-reaction methods
  • Explore the concept of oxidation states in more complex reactions
  • Practice with examples of redox reactions involving different compounds
USEFUL FOR

Chemistry students, educators, and anyone looking to deepen their understanding of redox reactions and oxidation states in chemical processes.

takando12
Messages
122
Reaction score
5

Homework Statement


[C2O4]-2 ------> CO2
Find if the reaction is oxidation or reduction and determine the no. of electrons lost or gained by carbon.

Homework Equations

The Attempt at a Solution


I started with finding the oxidation states of both elements. It's 3+ and 2- for carbon and oxygen on the LHS and 4+ and 2- for carbon and oxygen on the RHS. But i am having trouble in interpreting these values.
the OS of carbon goes from 3+ to 4+. My interpretation of this was that C had lost 3 electrons on the LHS while lost 4 electrons on the RHS to O. Hence I thought that C should have gained 1 electron. And this means it's reduction and that's the wrong answer. Can someone please correct my interpretation and help me out with this? This is quite a silly and basic question. Please bear with me, I am new to this.
 
Physics news on Phys.org
takando12 said:
C had lost 3 electrons on the LHS while lost 4 electrons on the RHS to O. Hence I thought that C should have gained 1 electron.

Think it over.

But in general, this is much easier done without oxidation numbers, just using half reactions. It is trivial to balance atoms, then it is trivial to see where the electrons have to be added (on which side) to balance the charge.
 
Borek said:
Think it over.

But in general, this is much easier done without oxidation numbers, just using half reactions. It is trivial to balance atoms, then it is trivial to see where the electrons have to be added (on which side) to balance the charge.
Ok I tried again. The OS of C on LHS is 3+ meaning there are 3 more protons than electrons. This goes to a 4+ on the RHS , 4 more protons than electrons , which means 1 electron must have been removed. And balancing the reaction we get 2CO2 on the LHS. So multiplying the OS with the no of atoms., i got 6+ on the LHS and 8+ on the RHS. Hence applying the same logic, 2 electrons must have been removed? Is this right?
 
Yes, electrons are removed.

C2O42- → 2CO2 + 2e-
 

Similar threads

Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
10K