A current of 2.25 A is applied to $\ce{NiCl2}$ solution

A. Write the balanced half reaction that takes place at the anode B. Write the balanced half reaction that takes place at the cathode

Can someone check if my answers seem logical? I'm not quite sure if I did the problem correctly

Anode: $\ce{O2 + 4H+ + 4e- -> 2H2O}$

Cathode: $\ce{2Ni^{2+} + 4e- -> 4Ni}$

  • $\begingroup$ Can you expand on why you think that answer should be correct i.e. why should the species selected by you get reduced/oxidised in the way you think they should? $\endgroup$ – stochastic13 Jan 10 '14 at 6:46
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    $\begingroup$ sure. when I looked at the standard reduction potential for the Nickel its value was -.25 V. When I looked at the Cl its value was 1.36. I knew that Ni was being reduced because because it's value was more positive than the -.83 of the standard reduction rection 2H20 + 2e- __> H2 + 2OH-. Then I looked at the Cl2 and compared it to the equaiton I had above for the anode. The reason I picked that as my answer is because the it had a standard cell potential of 1.23 and Cl2 was 1.32. Since 1.23 is LESS than Cl2, I chose it to be my answer for the oxidation half reaction $\endgroup$ – sloth1111 Jan 10 '14 at 14:09
  • $\begingroup$ You have written two reduction reactions. One of them needs to be an oxidation reaction. In a voltaic cell, where does the oxidation occur? $\endgroup$ – bobthechemist Jan 11 '14 at 23:38
  • $\begingroup$ the oxidation occurs at the anode $\endgroup$ – sloth1111 Jan 12 '14 at 18:22

At the A-node you O-xidize, at the C-athode you R-educe. (Vowel to vowel, consonant to consonant.)

Following this, you will have to generate electrons at the anode, which doesn't happen with your reaction.

What reaction will happen depends on the normal potential of each reaction, but I assume it will be something along the lines of $$\ce{2Cl- -> Cl2 ^ + 2e-} $$

Edit: I just noticed that the cathode reaction can be simplified to $$ \ce{Ni^{2+} + 2e- -> Ni} $$

  • $\begingroup$ ok thank you for showing me the proposed anode! I keep getting the reduction oxidation equations with the water and H20 mixed up with the actual cathode/anode equations. $\endgroup$ – sloth1111 Jan 9 '14 at 14:10

The reaction at cathode seems OK, but at the anode another reaction will occur.


At the cathode, 4 e$^-$ are consumed. The electrons have to be "generated" in the system.

Remark: The value of current applied is irrelevant in this kind of question. More relevant will be the applied potential.

  • $\begingroup$ Can you please explain to me what you mean by current? Is that the equation I initially put in for the anode. $\endgroup$ – sloth1111 Jan 9 '14 at 14:13
  • $\begingroup$ I meant the Electric current $\endgroup$ – Jaroslav Kotowski Jan 9 '14 at 19:56

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