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During the electrolysis of $\ce{Ni(NO3)2}$ using nickel electrodes, it's said that the molarity of the solution remains constant as nickel's active, and dissolves and deposits in equal measure.

But water has an oxidation potential of 0.828 V, as compared to nickel's 0.24. So while the cathodic deposition seems fine, shouldn't it be oxygen that's liberated at the anode?

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  • $\begingroup$ That's not how it works at all. Look again at that 0.828: what is the corresponding reaction? $\endgroup$ – Ivan Neretin Jul 1 at 14:21
  • $\begingroup$ Isn't it $2H_2O → O_2 + 4H^+ + 4e^−$, as opposed to $Ni → Ni^2+ + 2e^-$ $\endgroup$ – Harry Holmes Jul 1 at 14:32
  • $\begingroup$ At the cathode, the ion $\ce{Ni^{2+}}$ is reduced into metallic nickel. At the anode, the metallic nickel is more easily oxidized into $\ce{Ni^{2+}}$ ion, than water into $\ce{O_2}$. $\endgroup$ – Maurice Jul 1 at 19:05
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Some electrolysis experimets are best first performed and then an attempt at explanation.

For example, the products of the electrolysis of aqueous nickel nitrate with graphite electrodes are reportedly per a source (with video) as:

Electrolysis of a nickel nitrate solution produces oxygen at the anode, and hydrogen and nickel at the cathode.

Looking at related metal electrolysis, for example, Cu(NO3)2, per this source with again graphite electrodes:

Electrolysis of a copper(II) nitrate solution produces oxygen at the anode and copper at the cathode.

Lastly, with the transition metal manganese, a source on the electrolysis of Mn(NO3)2, reports, to quote:

$\ce{Mn(NO3)2 + 2H2O ± 2e- -> H2 + MnO2 + 2HNO3}$

Based on all of the above, and that hydrogen gas is forming per the reactions:

$\ce{H+ + e- -> .H }$

$\ce{.H + .H -> H2 (g)}$

I would expect similarly some transient nitric acid formation with Ni(NO3)2 per its potential interaction between the nitrate ion and the hydrogen atom radical as follows:

$\ce{ NO3- + .H -> .NO2 + OH-}$

$\ce{.NO2 + .NO2 <=> N2O4 }$

$\ce{N2O4 + H2O -> HNO2 + HNO3 }$

Note, my claimed introduction of HNO3 (and its likely subsequent removal by nickel metal) can explain the comment "the molarity of the solution remains constant as nickel's active, and dissolves and deposits in equal measure", and is also consistent with the electrolysis of Mn(NO3)2 resulting in the creation of nitric acid.

And lastly, you are correct, oxygen is produced at the anode.

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