Electrochemistry question

[BigLazy]

Advertisement - Members don't see this ad

Probably a stupid question, but then again, I'm pretty stupid to start studying one month ago.
When you are couplin oxidation with reduction for a galvanic cell. You pick the 1/2 rxn with the highest Standard reduction potential to be you reducing rxn. Consider the following example
Ni + 2e- --> Ni2+ E=0.5
Au+ 1e- --> Au+ E=-0.5

Do I have to multiple the Au reaction by a coefficient to get the 2 e- necessary for the reduction rxn?
and if yes, does the potential get doubled too?

thanks to whoever future doctor that answers

 

[Code Brown]

Probably a stupid question, but then again, I'm pretty stupid to start studying one month ago.
When you are couplin oxidation with reduction for a galvanic cell. You pick the 1/2 rxn with the highest Standard reduction potential to be you reducing rxn. Consider the following example
Ni + 2e- --> Ni2+ E=0.5
Au+ 1e- --> Au+ E=-0.5

Do I have to multiple the Au reaction by a coefficient to get the 2 e- necessary for the reduction rxn?
and if yes, does the potential get doubled too?

thanks to whoever future doctor that answers

No, your eV would still be 1.0. The # electrons is irrelevant when calculating eV. Good luck on saturday!

 

[Csv321]

You only multiply coefficients to balance equations and get a net equation. Reduction potentials do not get multiplied. They stay the same so if a passage gives you an unbalanced equation and asks for net potential, don't get thrown off by the "unbalanced" part. Just add up the reduction potentials as given.

Don't feel bad...I started exactly a month ago as well! 🙂

 

[BigLazy]

got it, so even though the rxn might be multiplied (useful when asked about the Q of the rxn), the potential stays the same when asked about voltage.

 

[VPDcurt]

Yes that is correct. It is different with enthalpy though - If I remember correctly. If you multiply the rxn by a number, you must do the same with the deltaH.