[ajax0604]

I have read quite a few resources and some say that H₂O is the competing species whereas others say that H⁺ and OH^- are the ones. I thought that perhaps it didn't matter which one you take, since H₂ is released regardless of whether you start with H₂O or H⁺. However, there was a question which asked what the expected products would be in the electrolysis of NiSO₄ solution. I thought that since H⁺ is the better oxidant, H₂ would be released but the answers say that Ni metal is deposited. This would make sense if Ni²⁺ was compared to H₂O since the former is the better oxidant according to the electrochemical series. Could someone clarify this?

[Borek]

Go for H⁺ and OH^-. If not for other reasons, concentration of H₂O is in most cases constant, while concentrations of its dissociation products vary greatly - and as they are involved in all water related electrode reactions, they can greatly change the potential at which the reaction takes place (think Nernst equation).

In the Ni²⁺/H⁺ case it all depends on the solution pH. The higher the pH, the lower the H⁺ reduction potential. At some point it gets lower that the reduction potential for Ni²⁺. To make things worse, at high pH you can expect hydroxide precipitation. For exact prediction of what would happen you need to consult Pourbaix diagram. If the question you mention didn't specify pH, it is impossible to answer - unless what the question really means is "what are the products you expect in the context of what you have been told when we discussed some simplified model of electrolysis".

[ajax0604]

Thank you, I think I understand the situation now. One more thing -you said that an increase in pH reduces the reduction potential of H⁺. Does this happen because there are more hydroxide ions in basic solutions, which makes the equilibrium shift to the left to release more H⁺ as per the equation 2H⁺ + 2e-    H₂? I'm afraid I don't know much about the Nernst equation as it is not in my course.

[Borek]

Potential depends on the concentration of H⁺ and pressure of the gaseous hydrogen.

[tex]E = 0.0 + \frac {RT}{2F} \ln \frac {[H^+]^2}{p_{H_2}}[/tex]

The former is expressed through pH.

[ajax0604]

I see. Based on the equation, if temperature increased, then E would also increase? If so, could you say that an increase in temperature makes H+ a better oxidant?

[Borek]

I see. Based on the equation, if temperature increased, then E would also increase?

Yes.

If so, could you say that an increase in temperature makes H+ a better oxidant?

Yes, but please remember that the potential for the reducing agent is a function of temperature as well.