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# emf Calculations watch

1. Hi
See in a summary questions (pg 55 AQA Chemistry (yr2) Oxford Ted Lister)

Calculate emf of 2H(+) + Pb ----> Pb (2+) + H2
E value Lead given as -0.13v and Hydrogen E value given as 0.00v (From a Table)

emf = E (red) - E (ox)
Pb 0-->2+ therefore loss 2 electrons (oxidized).
H +1 -->0 therefore gain 1 electon (reduced) E value given as 0.00v

emf = 0.00 - (-0.13)
emf =+ 0.13 v

However the answer in the back of the book gives 0.26v
Completed the other three questions similar to this one and got the right answer.
Is it me or the text book?
Many Thanks
OJ
2. (Original post by owjay)
Hi
See in a summary questions (pg 55 AQA Chemistry (yr2) Oxford Ted Lister)

Calculate emf of 2H(+) + Pb ----> Pb (2+) + H2
E value Lead given as -0.13v and Hydrogen E value given as 0.00v (From a Table)

emf = E (red) - E (ox)
Pb 0-->2+ therefore loss 2 electrons (oxidized).
H +1 -->0 therefore gain 1 electon (reduced) E value given as 0.00v

emf = 0.00 - (-0.13)
emf =+ 0.13 v

However the answer in the back of the book gives 0.26v
Completed the other three questions similar to this one and got the right answer.
Is it me or the text book?
Many Thanks
OJ
2H+ + 2e- ----> H2

Pb ----> Pb2+ + 2e-

Standard reduction potentials:

Reduction Eo = 0.00V

Reversed reaction

Pb2+ + 2e- ----> Pb

Eo = -0.13

Eooxidation = -Eoreduction

Eo oxidation = -(-0.13) = +0.13V

Eocell = Eoreduction + Eooxidation

Eocell = 0.00V + 0.13V = +0.13V
3. (Original post by owjay)
Hi
See in a summary questions (pg 55 AQA Chemistry (yr2) Oxford Ted Lister)

Calculate emf of 2H(+) + Pb ----> Pb (2+) + H2
E value Lead given as -0.13v and Hydrogen E value given as 0.00v (From a Table)

emf = E (red) - E (ox)
Pb 0-->2+ therefore loss 2 electrons (oxidized).
H +1 -->0 therefore gain 1 electon (reduced) E value given as 0.00v

emf = 0.00 - (-0.13)
emf =+ 0.13 v

However the answer in the back of the book gives 0.26v
Completed the other three questions similar to this one and got the right answer.
Is it me or the text book?
Many Thanks
OJ
From everything I know about electrochemistry (Currently a bit, doing a masters module in it), the answer should be +0.13V
You shouldn't multiply the E value by the number of electrons in the equation (Nernst equation proves this)
I don't know if you are taught differently at A level or if the text book is wrong (Mine was notorious for mistakes, I spent two years just making corrections)

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