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# Chemistry- calculating E cell watch

1. a bit confused atm

so

Zn2+(aq) + 2e ---> Zn(s) E= - 0.76V
Fe2+(aq) + 2e ---> Fe(s) E= -0.44V

From the equation, they're both being reduced.. Sooo how can I calculate Ecell? On chem revise its saying most negative=oxidised going from right to left...

But its getting reduced, not oxidised

Thanks lol
2. Feraligatr
James A
C0balt

I think you're all offline atm too lol
3. bumpppppp
4. Nvmmmmmm I get it
5. Simple: you reverse the one that would create the highest (most positive) V overall. So, Zn2 would become an oxidation.
6. (Original post by Dinaa)
a bit confused atm

so

Zn2+(aq) + 2e ---> Zn(s) E= - 0.76V
Fe2+(aq) + 2e ---> Fe(s) E= -0.44V

From the equation, they're both being reduced.. Sooo how can I calculate Ecell? On chem revise its saying most negative=oxidised going from right to left...

But its getting reduced, not oxidised

Thanks lol
Personally I just maximise the E cell value, so in order to maximise the total E cell I want the Fe2+ one to be positive rather than negative so it goes the other way, so overall my equation is Fe(s) +Zn2+(aq) ---> Zn(s) +Fe2+(aq) and the E cell is 0.76+-(-0.44) =1.1V

edit: oops didn't see - in front of Zn Ecell value, so that would go the other way instead as it is more negative, and Fe would stay the same way as it is in data book
7. I am recently quite useless at replying quickly just so you know...rarely on TSR on laptop these days (my living space is quite dead on here you see, no fun&busy mega thread or offer holder thread)
8. (Original post by Dinaa)
a bit confused atm

so

Zn2+(aq) + 2e ---> Zn(s) E= - 0.76V
Fe2+(aq) + 2e ---> Fe(s) E= -0.44V

From the equation, they're both being reduced.. Sooo how can I calculate Ecell? On chem revise its saying most negative=oxidised going from right to left...

But its getting reduced, not oxidised

Thanks lol
I haven't been using TSR as much
My Chem teacher who teaches this taught this slightly differently to Chemrevise

You flip over the more negative one making that one positive (both are reduced, you need one which is oxidised and one which is reduced)

Zn2+(aq) + 2e ---> Zn(s) E= - 0.76V
Fe2+(aq) + 2e ---> Fe(s) E= -0.44V

Zn half-equation is more negative and therefore

Zn(s) --->Zn2+(aq) + 2e E= +0.76V
Fe2+(aq) + 2e ---> Fe(s) E= -0.44V

Add and balance (you don't need to balance here really)

Zn(s) +Fe2+(aq) + 2e --->Zn2+(aq) + 2e + Fe(s)
+0.76 + -0.44 = +0.32
9. (Original post by Feraligatr)
I haven't been using TSR as much
My Chem teacher who teaches this taught this slightly differently to Chemrevise

You flip over the more negative one making that one positive (both are reduced, you need one which is oxidised and one which is reduced)

Zn2+(aq) + 2e ---> Zn(s) E= - 0.76V
Fe2+(aq) + 2e ---> Fe(s) E= -0.44V

Zn half-equation is more negative and therefore

Zn(s) --->Zn2+(aq) + 2e E= +0.76V
Fe2+(aq) + 2e ---> Fe(s) E= -0.44V

Add and balance (you don't need to balance here really)

Zn(s) +Fe2+(aq) + 2e --->Zn2+(aq) + 2e + Fe(s)
+0.76 + -0.44 = +0.32
Is this always the case?

Thank you
10. (Original post by Dinaa)
Is this always the case?

Thank you
Yes you always flip the most negative one and make it positive, unless you were trying to find out if a reaction was possible in which case you need to do whatever reaction is stated in the question
11. (Original post by samb1234)
Yes you always flip the most negative one and make it positive, unless you were trying to find out if a reaction was possible in which case you need to do whatever reaction is stated in the question
prsom, thank you
12. (Original post by Dinaa)
Is this always the case?

Thank you
Sorry I just saw this
You're welcome
And it is

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