6O.1

Moderators: Chem_Mod, Chem_Admin

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

6O.1

Postby Alicia Lin 2F » Sat Feb 29, 2020 2:32 pm

In the solution manual, it says to choose the reduction reaction with most positive standard reduction potential for the cathode and choose reduction reaction with most negative standard reduction potential for the anode. But they choose Ni2+ reaction for cathode which has E=-0.23V and choose H2O reaction for anode which has E=+1.23V. Using their reasoning described, shouldn't the cathode reaction be the H2O reaction since it is the most positive standard reduction?

Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

Re: 6O.1

Postby Chem_Mod » Sat Feb 29, 2020 9:42 pm

The reduction half reaction you should be considering for water is 2H2O + 2e —> H2 + 2OH- with Ecell= -0.83 V
The half reaction you are thinking of is for the reduction of O2 into water

AlyssaYeh_1B
Posts: 100
Joined: Sat Aug 17, 2019 12:16 am

Re: 6O.1

Postby AlyssaYeh_1B » Sun Mar 01, 2020 4:04 pm

If the standard reduction potentials that we are meant to use are -0.23V and -0.83V, how is the minimum potential that must be supplied for the onset of electrolysis +1.46V?

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

Re: 6O.1

Postby DHavo_1E » Thu Mar 12, 2020 6:04 am

Chem_Mod wrote:The reduction half reaction you should be considering for water is 2H2O + 2e —> H2 + 2OH- with Ecell= -0.83 V
The half reaction you are thinking of is for the reduction of O2 into water


Hello,

I am also a bit confused on this question because the book chooses the cathode as Ni2+(aq) +2e- ->Ni(s) instead of the reduction half reaction for water with Ecell = -0.83V, and the anode as the oxidation of water 2H2P(l)-> O2(g) = 4H+ (aq) +4e- with Ecell= +1.23V. Why is that?


Return to “Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust”

Who is online

Users browsing this forum: No registered users and 4 guests