If this is the reaction for a electrolytic cell being charged:
2Ni(OH)2 (s) + Zn(OH)2(s) ---> 2Ni(OH)(s) + Zn(s) +2H2O(l)
How do I write the half equation for the negative electrode as the cell is being recharged and the half equation of the positive electrode as the cell is being discharged? How do I know which one is which?
When a cell is being recharged, it's practically electrolysis where the negative electrode is the cathode. This means reduction will occur here. Look at the oxidation states of your RECHARGING reaction above and note that the Zn goes from +2 to 0 so a reduction, which will occur at the negative cathode in the recharge.
2e- Zn(OH)2 ---> Zn + 2OH-
The positive electrode during DISCHARGE is the cathode like in a galvanic cell. Here there is reduction (as always with a cathode) BUT be careful! This is discharge so you have to look at the reaction backwards and not what is being reduced??
Did you write out the equation correctly? I think the Ni(OH)(s) on the right hand side of your equation is meant to be NiO(OH) from memory. I'm not sure if it balances.
2NiOOH + Zn + 2H2O -> 2Ni(OH)2 + Zn(OH)2 This is what I found in a practice exam example of this discharge reaction, so I'll go from here for explanation purposes. Ni oxidation state goes from +3 to +2 hence a reduction.
2e + 2NiOOH + 2H2O ---> 2Ni(OH)2 + 2OH- you remove the NiOOH and Ni(OH)2 onto a separate line, then balance as per normal for a redox. When you need to use H+ as per an acidic environment, you add alkaline OH- to both side to get rid of the H+ as you can tell from all the OH- everywhere in the full thing, that this is in an alkaline environment.
Home
Help
Search
Login
