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Standard Electrode Potentials

We weren't really taught this well so now I've come to revise it and I'm failing miserably.
I'm trying to do this worksheet:
uploadfromtaptalk1365089064843.jpg

so far I've got:
1) zero; as it can be used to compare other reactions to see if they are a reduction or oxidation

2) Zn (s) + Cd^2+ (aq) -> Zn^2+ (aq) + Cd (s)

3) -0.4--0.76=0.36v (is this right?)

4) I'm stuck. Is it just -0.34--0.26? It asks about reducing Cu^2- to copper but the table shows values for Cu^2+ :dontknow:

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Original post by Squirb
We weren't really taught this well so now I've come to revise it and I'm failing miserably.
I'm trying to do this worksheet:
uploadfromtaptalk1365089064843.jpg

so far I've got:
1) zero; as it can be used to compare other reactions to see if they are a reduction or oxidation

correct


2) Zn (s) + Cd^2+ (aq) -> Zn^2+ (aq) + Cd (s)

correct


3) -0.4--0.76=0.36v (is this right?)

correct


4) I'm stuck. Is it just -0.34--0.26? It asks about reducing Cu^2- to copper but the table shows values for Cu^2+ :dontknow:

Posted from TSR Mobile


If the reaction were feasible vanadium(II) would be the oxidised state and Cu(II) the reduced state.

E = E(red)-E(ox) = +0.34 -- 0.26 = +0.60 V

This is positive and greater then 0.3V, so the reaction is spontaneous (feasible) and goes to completion...
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Avatar for Squirb
Squirb
OP
15
Original post by charco

If the reaction were feasible vanadium(II) would be the oxidised state and Cu(II) the reduced state.

E = E(red)-E(ox) = +0.34 -- 0.26 = +0.60 V

This is positive and greater then 0.3V, so the reaction is spontaneous (feasible) and goes to completion...


Thank you very much! I think I was just over-complicating the last one:colondollar:

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