14.13 (d)

Moderators: Chem_Mod, Chem_Admin

Hannah Guo 3D
Posts: 56
Joined: Fri Sep 29, 2017 7:06 am

14.13 (d)

Postby Hannah Guo 3D » Sun Feb 18, 2018 5:50 pm

Based on the original chemical equation, Au^+ was reduced to Au(s) and was oxidized to Au^(3+), but the answer reverses the anode reaction and yield Au(s)-->Au^(3+) + 3e^-. Can someone please explain why we need to reverse the half-reaction? Thank you!

Srbui Azarapetian 2C
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 2 times

Re: 14.13 (d)

Postby Srbui Azarapetian 2C » Sun Feb 18, 2018 8:47 pm

The equation must be reversed because the original oxidation reaction is Au3+(aq) + 3e- -> Au (s). This must be reversed into Au (s) -> Au3+(aq) + 3e- so that the Au 3+ ends up in the products side of the cell reaction given in the problem, Au3+ (aq) -> Au(s) + Au3+ (aq).

Lindsay H 2B
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Re: 14.13 (d)

Postby Lindsay H 2B » Sun Feb 18, 2018 9:56 pm

I'm stuck on this question too. For the oxidation half-reaction, why is it Au(s)--> Au^3+(aq) + 3e- instead of Au+(aq)--> Au^3+(aq) + 2e-? Since the original reaction makes it seem like Au+(aq) is getting oxidized, not Au(s).

Return to “Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams”

Who is online

Users browsing this forum: No registered users and 1 guest