Standard Reduction Potential
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Standard Reduction Potential
How come the standard reduction potentials stay the same when half reactions are balanced?
Re: Standard Reduction Potential
I think because standard reduction potential is an intensive property so the amount that the reaction occurs doesn't matter.
Re: Standard Reduction Potential
From a mathematical standpoint, if you take a look at the equation deltaG = -nFE, and rearrange it into E = -deltaG/nF, you will see that if we multiply the chemical reaction by a factor, both deltaG and n change by that factor, so the net result is no change to the value of E.
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Re: Standard Reduction Potential
Even when you multiply the half reactions by a factor in order to balance the number of electrons, the overall change of E does not change, while the value of delta G does.
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