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E0 calculations (1 Viewer)

currysauce

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i need help

i am not understanding this

in this revision sheet... it has a question

fe ---> fe2+ +2e- 0.44V <------ but in the book... it has - 0.44V

what does this mean?

is this because they are, in the sheet, using the oxidation potential thus i should change the sign?
 

~*HSC 4 life*~

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normally oxidation of iron
ie

Fe(s) --> Fe(2+) + 2e(-) E(oxidation)= 0.44V

im not sure what the qu is, or what the book is showing you, but if you look at a normal table of standard of electrode potentials they all show you reduction potentials, so if you want an oxidation potential you reverse the sign.

so i guess the last qu of ur post is correct, you reverse the sign because the E(reduction) given in a table is -0.44V, but you want oxidation which does not require an imput of power supply, it will automatically give up its electrons and produce 0.44V energy
 

tabularasa

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Well the standard reduction table shows the ions and electrons truning into the solid. So noramlkly this is a negative number so when you rever it to get Fe into Fe2+ you reverse the equation adn thus it has to switch the sign in front of the voltage.
 

jamesy_1988

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just make sure that you reverse the sign if it is an oxidation reaction and you will get it right.

Jamesy
 

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