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# Standard Electrode Potentials watch

1. 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:

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+

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2. (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:

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+

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...
3. (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

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Updated: April 4, 2013
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