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# help on this question? electrochemical cells. watch

1. can someone calculate step by step the emf of this?

I2 + 2Br- ---> Br2 + 2I-

I2 + 2e- ---> 2I- =+0.54V
Br2 + 2e- ---> 2Br- = +1.07V

Its question 2b page 207 on NT book btw

thanks.
2. (Original post by JSN)
can someone calculate step by step the emf of this?

I2 + 2Br- ---> Br2 + 2I-

I2 + 2e- ---> 2I- =+0.54V
Br2 + 2e- ---> 2Br- = +1.07V

Its question 2b page 207 on NT book btw

thanks.
Put the half equations the right way to show reduction and oxidation. A redox reaction cannot have both reactions as oxidising, one must reduce and one must oxidise!
3. (Original post by Booyah)
Put the half equations the right way to show reduction and oxidation. A redox reaction cannot have both reactions as oxidising, one must reduce and one must oxidise!
would it be..

2Br- ---> Br2 + 2e- = +1.07V
I2 + 2e- ---> 2I- = -0.54V

?
4. (Original post by JSN)
would it be..

2Br- ---> Br2 + 2e- = +1.07V
I2 + 2e- ---> 2I- = -0.54V

?
Exactly, and as easy as that was all you have to do is add the values, if they are above o then the reaction is feasible
5. (Original post by Booyah)
Exactly, and as easy as that was all you have to do is add the values, if they are above o then the reaction is feasible
The answer im getting is +0.53V, the textbook says the answer is -0.53V where am I going wrong do you think?

Thanks for the help..
6. (Original post by JSN)
The answer im getting is +0.53V, the textbook says the answer is -0.53V where am I going wrong do you think?

Thanks for the help..
I'm an idiot, forgive me for telling you you were correct,
The full equation is:

I2 + 2Br- ---> Br2 + 2I-

The half equations are:

I2+ 2e- ----> 2I- = +0.54
and

2Br- ----> Br2 + 2e- = -1.07

Which gives you -0.53

Sorry for the confusion, I misread the half equations earlier.
7. (Original post by Booyah)
I'm an idiot, forgive me for telling you you were correct,
The full equation is:

I2 + 2Br- ---> Br2 + 2I-

The half equations are:

I2+ 2e- ----> 2I- = +0.54
and

2Br- ----> Br2 + 2e- = -1.07

Which gives you -0.53

Sorry for the confusion, I misread the half equations earlier.
thank you so much, cleared up alot there!
appreciate it REPPED!
8. (Original post by JSN)
thank you so much, cleared up alot there!
appreciate it REPPED!
No problem
9. (Original post by Booyah)
No problem
last question why would the answer for this be +0.26v not +0.13?!

2H+ + Pb ---> Pb2+ + H2

2H+ + 2e- ---> H2 = 0.00V
Pb2+ + 2e- ---> Pb =-0.13V

therefore

Pb ---> Pb2+ + 2e- = +0.13V
10. (Original post by JSN)
last question why would the answer for this be +0.26v not +0.13?!

2H+ + Pb ---> Pb2+ + H2

2H+ + 2e- ---> H2 = 0.00V
Pb2+ + 2e- ---> Pb =-0.13V

therefore

Pb ---> Pb2+ + 2e- = +0.13V
I'm gonna be honest and say I do not know... I am sure +0.13 would be correct, but check it out with your teacher, he or she can probably tell you why, and post it on here I would like to know why it is right!
11. (Original post by Booyah)
I'm gonna be honest and say I do not know... I am sure +0.13 would be correct, but check it out with your teacher, he or she can probably tell you why, and post it on here I would like to know why it is right!
Haha no worries and I will do tomorrow and I will post it on here

and for the first question what would the overall sum be to get the answer just do know what to minus together?
12. (Original post by JSN)
Haha no worries and I will do tomorrow and I will post it on here

and for the first question what would the overall sum be to get the answer just do know what to minus together?
Well I have never used an equation, I tend to just add the values as one will be reducing and the other oxidising, I don't know any equation as such...

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