34 Replies
@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.if you see the change in elemental oxidation state of Mn throughout the reaction, on LHS it is in +7 state, and on the RHS it went to +2 state, so net change in its oxidation state is 7-2=+5
thats the net electron exchange for the reaction as well
Why can't we just subtract net charges for net e change
you are doing +2 - (-1)?
No
you can on half reactions but they oxn and red halves should be perfectly balanced then....
Net charge on lhs = 2
On rhs = 5
e transfer = 3
bro see its a redox reaction
the participating elemental metal atoms (Mn and Fe) undergo reduction and oxn respectively
Ohhh in that case why did u chose to see Mn's oxid only
bro first you need to see that the eqn isnt balanced
so first you need to get it into the balanced form....
Ok
did you get it? has your doubt been resolved?
you can see elemental oxn changes after balancing......
whats with such high count on slow mode? ☠️
Ikr
So bal eqn is
8H+ + 5Fe2+ + MnO4- = 5Fe3+ + Mn2+ + 4H2O
you can see yourself now whether you use Fe now or Mn 😄
How? I don't get it
you can use both now, for Fe elemental change is 15-10=5 and for mn 7-2=5
Ohhhhhhh ok I get it
way to go buddy 👍 all the best 🙏
Wait for both elemental change is 5
yup so your n=5
So net e transfer shouldn't it be 10 5 for each?
☠️ bro see one of them is "giving" these e- and the other is taking them....you will be counting the same thing 2 times.
Oh mb 💀
you can see the criss cross method of balancing redox reactions once it will give you a nice idea bout e- transfers through oxidation state changes....
Thx I got it now
also if you are doing electrochem i would suggest to have a read on the thermodynamics of cell part....i believe you would have already done it but i have seen some places its not mentioned so.....
What is that
Like G and S in terms of Ecell?
and more....like the coefficient (thermal) of a cell.....
Ye I have done that
i think you would have done it.....already...if not you can have a read....it would be there in coaching materials in theory part i believe.
ok 🙏 all the best 👍
Thx u very much
+solved @Deleted User
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