The Student Room Group

This discussion is now closed.

Check out other Related discussions

REDOX- electrode potential calculations help please!

I'm attempting my chemistry homework and getting nowhere with these electrode potential calculations :emo:

What i don't get is what values we use to get the standard EMF.(i know that use the standard potential reduction table-but how do i know whether to use 2H20 + 2e- ---> H2(g) + 2OH- orrr 2 H+(aq) + 2 e- -----> H2(g) ) ??

+ how do i know which element/solution is the one being oxidised and reduced??

This is the example i'm stuck on atm:

Pt(s) | H2(g)|2H+(aq) || Br2(l)|2Br-(aq)|Pt(s)

I'm very much confused by this (missed the crucial lesson due to exams last week :sad: )

If anyone could explain this to me, i'd really appreciate it! :h: (+rep if its really of that much value!:grin: )

Thank you :smile:
Reply 1
Avatar for time.to.dance
time.to.dance
OP
8
Anyone?? :sad:
time.to.dance
I'm attempting my chemistry homework and getting nowhere with these electrode potential calculations :emo:

What i don't get is what values we use to get the standard EMF.(i know that use the standard potential reduction table-but how do i know whether to use 2H20 + 2e- ---> H2(g) + 2OH- orrr 2 H+(aq) + 2 e- -----> H2(g) ) ??

You alwsy use the hydrogen half equation unless specifically dealing iwth a base.


+ how do i know which element/solution is the one being oxidised and reduced??

This is the example i'm stuck on atm:

Pt(s) | H2(g)|2H+(aq) || Br2(l)|2Br-(aq)|Pt(s)


In the case of cell reactions the electrons can only go one way and that is from the most reactive reducing agent (i.e. the substance with the largest negative electrode potential) to the most reactive oxidising agent (i.e. the one with the largest positive electrode potential)

In your cell above the two half reactions (both traditionally written aqs reductions) are:

H+ + 1e --> 1/2H2
1/2Br2 + 1e --> Br-

Of these two the bromine half equation is the most positive and the hydrogen the most negative (although it is zero as the standard)

Hence the electrons must flow from the hydrogen half cell to the bromine half cell. For the hydrogen to release electrons the half equation must be going backwards:

1/2H2 --> H+ + 1e

and the bromine forwards:

1/2Br2 + 1e --> Br-

Hence the overall cell reaction is obtained by adding them together (the electrons are already equal)

1/2H2 + 1/2Br2 -> H+ + Br-
Pt(s) | H2(g)|2H+(aq) || Br2(l)|2Br-(aq)|Pt(s)

| = phase boundary

|| = Salt bridge

Left hand side (LHS) = oxidation => H2(g) ---> 2H+ + 2e-

RHS = reduction => Br2 + 2e- ---> 2Br-

edit : slow :rolleyes:
Reply 4
Avatar for time.to.dance
time.to.dance
OP
8
charco
.

EierVonSatan
.

Thank you both :top:
It is much clearer now and helped me with my work. :biggrin:
will rep when i can :h:

Related discussions

Latest

Trending

Trending

Articles for you