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    Hi,

    I have a question about calculating the cell e.m.f.'s.

    Say you had a cell diagram as such:

    Ni (s) | Ni2+ (aq) | Zn2+ (aq) | Zn (s)

    So I am trying to calculate the cell e.m.f. of this cell, and then write the equation that would occur if the cell were short-circuited.

    The data I have (as reduction potentials) is:

    Ni2+ (aq) + 2e- ------> Ni (s) ______Ecell = -0.25V
    Zn2+ (aq) + 2e- ------> Zn (s) _____Ecell = -0.76V

    Can I just switch the sign of the top equation to make it positive (i.e. +0.25V) in order to work out the Ecell value when Ni is oxidised to Ni2+ and 2e-?

    This would make my e.m.f. = -0.76 + 0.25 = -0.51V

    Meaning my overall equation if the cell were short-circuited is:
    Ni2+ + Zn ----> Ni + Zn2+

    Is this correct?

    Thanks for your help
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    (Original post by Funky_Giraffe)
    Hi,

    I have a question about calculating the cell e.m.f.'s.

    Say you had a cell diagram as such:

    Ni (s) | Ni2+ (aq) | Zn2+ (aq) | Zn (s)

    So I am trying to calculate the cell e.m.f. of this cell, and then write the equation that would occur if the cell were short-circuited.

    The data I have (as reduction potentials) is:

    Ni2+ (aq) + 2e- ------> Ni (s) ______Ecell = -0.25V
    Zn2+ (aq) + 2e- ------> Zn (s) _____Ecell = -0.76V

    Can I just switch the sign of the top equation to make it positive (i.e. +0.25V) in order to work out the Ecell value when Ni is oxidised to Ni2+ and 2e-?

    This would make my e.m.f. = -0.76 + 0.25 = -0.51V

    Meaning my overall equation if the cell were short-circuited is:
    Ni2+ + Zn ----> Ni + Zn2+

    Is this correct?

    Thanks for your help
    Firstly, the zinc half-cell should be on the left since it has a lower Eo value. To find the emf of a cell, all you have to do is take the more negative E value from the less negative E value.

    Ecell = Eless -ve - Eless -ve

    Using this method you will always get the emf value, which is positive. However, if you did this the other way around and got a negative answer all you have to do is take the positive value. Emf is always positive.

    The half-cell that has the more negative electrode potential acts as the reducing agent. That means that in this case, zinc acts as the reducing agent and hence gets oxidised itself. Therefore, yes, the overall equation which you wrote is correct.
 
 
 
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