[Zishi]

I'm facing problem with part(b) of this question:


Here are the Electrode potential values of half cells that are being used:


There are many ways of finding the Electrode potential of an electrochemical cell, by using different formulas. But I prefer to add up the electrode potentials of two half cells after reversing the sign of the electrode potential of the half cell that is shifting towards the left hand side of the equation. The equilibrium with more positive E naught value shifts to the right and and less positive(or more negative) one shifts to the left.

So if I apply the same concept here, the reactions are occurring in opposite direction to what E naught values predict, so I get negative Cell electrode potential values. Mark scheme just subtracts the electrode potential value of reaction at cathode from electrode potential values of reactions at anode.

Where am I going wrong?

[aeyurttaser13]

(Original post by Zishi)
I'm facing problem with part(b) of this question:


Here are the Electrode potential values of half cells that are being used:


There are many ways of finding the Electrode potential of an electrochemical cell, by using different formulas. But I prefer to add up the electrode potentials of two half cells after reversing the sign of the electrode potential of the half cell that is shifting towards the left hand side of the equation. The equilibrium with more positive E naught value shifts to the right and and less positive(or more negative) one shifts to the left.

So if I apply the same concept here, the reactions are occurring in opposite direction to what E naught values predict, so I get negative Cell electrode potential values. Mark scheme just subtracts the electrode potential value of reaction at cathode from electrode potential values of reactions at anode.

Where am I going wrong?

the way i do it which is right is, always take the more negative value, flip it, and add it to the other value.
so if u have potentials of -0.8 and -0.3, the more negative one is 0.8, so flip it, and add it to the other value and ull get 0.5. that is the spontaneous reaction. if u dont get a positive value when u do this, then it wouldn't be spontaneous.

[Zishi]

(Original post by aeyurttaser13)
the way i do it which is right is, always take the more negative value, flip it, and add it to the other value.
so if u have potentials of -0.8 and -0.3, the more negative one is 0.8, so flip it, and add it to the other value and ull get 0.5. that is the spontaneous reaction. if u dont get a positive value when u do this, then it wouldn't be spontaneous.

I would like to get a second opinion on it.

BUMP!

[EierVonSatan]

My electrochemistry is so rusty I wouldn't put too much faith on me being right here I'd rather drink off milk than do electrochem...

E_cell = E(more positive) - E(less positive) (NB when they're written as reductions)

Also in question a you never write negative quantities in chemical equations so should be 2H2O ---> 4H+ + O2 + 4e- instead

[Zishi]

(Original post by EierVonSatan)
My electrochemistry is so rusty I wouldn't put too much faith on me being right here I'd rather drink off milk than do electrochem...

E_cell = E(more positive) - E(less positive) (NB when they're written as reductions)

Also in question a you never write negative quantities in chemical equations so should be 2H2O ---> 4H+ + O2 + 4e- instead

Ah, right. So the "equilibrium with more positive E^o value will move to right and the equilibrium with less positive E^o will move to left" rule won't apply here because that's an electrolytic cell, not an electrochemical cell, right?

[EierVonSatan]

(Original post by Zishi)
Ah, right. So the "equilibrium with more positive E^o value will move to right and the equilibrium with less positive E^o will move to left" rule won't apply here because that's an electrolytic cell, not an electrochemical cell, right?

Moving right/left means nothing to me, sorry

It's probably better you stick to the way you've been taught to do this...

[Zishi]

(Original post by EierVonSatan)
Moving right/left means nothing to me, sorry

It's probably better you stick to the way you've been taught to do this...

Ahh, no problem. Thanks anyway.

I hope charco will help.

(Original post by charco)
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