[Darren]

What are the factors that affect the value for the electrode potential for half equations? Take for example the equation of Cu²⁺+2e^- Cu

I know that if you add sodium hydroxide to this equilibrium, the standard electrode potential will decrease as by Le Chatelier's principle, the equilibrium position shifts to the left to increase the concentration of Cu²⁺ ions.

But how does temperature affect the standard electrode potential?

How does the voltage applied to a solution containing the species present in the equilibrium of the half equation affect it also? Since the electrode potential is measured in volts, or potential, does voltage being APPLIED to it cause any change to the equilibrium or to the standard electrode potential value?

What other factors are there that affect the equilibrium position and also for the electrode potential value?

[Borek]

http://en.wikipedia.org/wiki/Nernst_equation

[Darren]

http://en.wikipedia.org/wiki/Nernst_equation

Could you simplify the whole idea for me please? I have never learnt that before and i cant understand the mathematical formulae but i would like to know the general idea, in qualitative form if possible so that i can understand it almost completely in words

[Borek]

I will give you an example. For the half reaction of permanganate reduction to Mn²⁺ (or Mn²⁺ oxidation to permanganate - they occur at the same potential)

MnO₄^- + 8H⁺ + 5e^-   Mn²⁺ + 4H₂O

potential is given by

[tex]E=E_0 + \frac {RT}{5F} \ln \frac{[MnO_4^-][H^+]^8}{[Mn^{2+}]}[/tex]

where E₀ is so called standard potential, measured and put in tables.

[Darren]

I will give you an example. For the half reaction of permanganate reduction to Mn²⁺ (or Mn²⁺ oxidation to permanganate - they occur at the same potential)

MnO₄^- + 8H⁺ + 5e^-  Mn²⁺ + 4H₂O

potential is given by

[tex]E=E_0 + \frac {RT}{5F} \ln \frac{[MnO_4^-][H^+]^8}{[Mn^{2+}]}[/tex]

where E₀ is so called standard potential, measured and put in tables.

If the potential applied is increased above the potential for that half equation, can i say that the forward reaction, reduction, will take place more readily than the oxidation reaction?

[Borek]

If you are forcing the potential onto the system, system will react till its own potential will be identical to the forced one. So if you increase the potential you can expect oxidation of Mn²⁺ to permanganate.

[Darren]

If you are forcing the potential onto the system, system will react till its own potential will be identical to the forced one. So if you increase the potential you can expect oxidation of Mn²⁺ to permanganate.

Ooh! So that means this is also like le chatelier's principle? When you add the voltage, the equilibrium will shift so that the voltage it produced is altered to counteract the change of vontage you applied to it?

[Borek]

In a way... But you can also see half cell this way: when you apply a different potential to the electrode you are forcing the system to react till its potential equals that of the electrode.