Balance in an acidic solution redox

[nautica]

Check this

Balance in an acidic solution

Fe2+ + MnO4- ----- Fe3+ + Mn2+

(Fe2+ ---- Fe3+ + 1e-) x 7
7e- + 8H+ + MnO4- ----- Mn + 4H2O

8H+ + 7Fe2+ + MnO4- ----- 7Fe3+ + Mn + 4H2O

Thanks
Nautica

 

[chem_tr]

I have not gone deep in your post; there are two strict ways to check if a redox reaction is okay or not:

1. Check for a balance for atoms in two sides.
2. Check for a balance for ions (electron count) in two sides

If they are okay, your reaction is correct.

 

[ravenprp]

Thank you for sharing the redox reaction and asking for assistance in balancing it in an acidic solution. The key to balancing redox reactions is to make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. In this case, we can see that Fe2+ is being oxidized to Fe3+, which means it is losing one electron. On the other hand, MnO4- is being reduced to Mn2+, which means it is gaining five electrons. Therefore, we need to multiply the Fe2+ half-reaction by 7 and the MnO4- half-reaction by 5 to balance the number of electrons on both sides.

Next, we need to balance the number of hydrogen ions (H+) and water molecules (H2O) on both sides of the equation. In an acidic solution, we can add H+ ions to balance the hydrogen atoms and H2O molecules to balance the oxygen atoms. In this case, we need to add 8 H+ ions to the Fe2+ half-reaction and 4 H2O molecules to the MnO4- half-reaction.

Finally, we can combine the two half-reactions to get the balanced equation:

8H+ + 7Fe2+ + 5MnO4- → 7Fe3+ + Mn2+ + 4H2O

This equation is now balanced in an acidic solution. I hope this helps! Please let me know if you have any further questions.