ROadmanKys
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The following cell has an EMF of +0.46 V.

Cu Cu2+ Ag+ Ag

Which statement is correct about the operation of the cell?

A Metallic copper is oxidised by Ag+ ions.

B The silver electrode has a negative polarity.

C The silver electrode gradually dissolves to form Ag+ ions.

D Electrons flow from the silver electrode to the copper electrode
via an external circuit.

(Total 1 mark)
how would you know its A?
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charco
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(Original post by ROadmanKys)
The following cell has an EMF of +0.46 V.

Cu Cu2+ Ag+ Ag

Which statement is correct about the operation of the cell?

A Metallic copper is oxidised by Ag+ ions.

B The silver electrode has a negative polarity.

C The silver electrode gradually dissolves to form Ag+ ions.

D Electrons flow from the silver electrode to the copper electrode
via an external circuit.

(Total 1 mark)
how would you know its A?
Convention shows the negative electrode on the left and positive on the right.
Hence, electrons flow around the external circuit from left to right.
Oxidation occurs on the left hand side and reduction on the right hand side.
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ROadmanKys
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(Original post by charco)
Convention shows the negative electrode on the left and positive on the right.
Hence, electrons flow around the external circuit from left to right.
Oxidation occurs on the left hand side and reduction on the right hand side.
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Size:  85.3 KBfor (C) how would i determine that its cobalt3+ and not cobalt 2+ ? is there a special condition for reacting with water? if its with Co3+, then h20 would act as a reducing agent since it gets oxidised.
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charco
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(Original post by ROadmanKys)
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Size:  85.3 KBfor (C) how would i determine that its cobalt3+ and not cobalt 2+ ? is there a special condition for reacting with water? if its with Co3+, then h20 would act as a reducing agent since it gets oxidised.
You have to compare the half equation given with water with the two cobalt half equations, using E(cell) = E(red) - E(ox) . A spontaneous reaction will have E(cell) positive and greater than +0.3V.
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ROadmanKys
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(Original post by charco)
You have to compare the half equation given with water with the two cobalt half equations, using E(cell) = E(red) - E(ox) . A spontaneous reaction will have E(cell) positive and greater than +0.3V.
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in this case, why cant fe2+ be used as it can be oxidised, and the h20 can be reduced? or is there just a fact that when theres a question like this that arises, chose the E- with the highest/most + value?
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charco
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(Original post by ROadmanKys)
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in this case, why cant fe2+ be used as it can be oxidised, and the h20 can be reduced? or is there just a fact that when theres a question like this that arises, chose the E- with the highest/most + value?
There are no half-equations for water in this example ... !
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ROadmanKys
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i know, but it gave two options of Fe, fe2+ and fe3+ but it used fe3+ why? also you said something about being spontaneous and a reaction wont occur without it being one. but how do i determine that it is spontaneous?
(Original post by charco)
There are no half-equations for water in this example ... !
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charco
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(Original post by ROadmanKys)
i know, but it gave two options of Fe, fe2+ and fe3+ but it used fe3+ why? also you said something about being spontaneous and a reaction wont occur without it being one. but how do i determine that it is spontaneous?
If you read my earlier post I said "You have to compare the half equation given with water with the two cobalt half equations, using E(cell) = E(red) - E(ox) . A spontaneous reaction will have E(cell) positive and greater than +0.3V"

That's how you determine whether or not a reaction is thermodynamically spontaneous (using electrode potentials)

The hydrogen half equation has the electrode potential = 0.0V
The iron(II)|Fe half equation has an electrode potential = -0.44

Hence, for the process:

Fe2+ + H2 --> Fe + 2H+

Using E(cell) = E(red) - E(ox)

E(cell) = -0.44 - 0 = -0.44V

This is negative and therefore the reaction will not happen, it is not spontaneous.

However for the reaction:

2Fe3+ + H2 --> 2Fe2+ + 2H+

E(cell) = +0.77 - 0.0 = +0.77

This reaction is spontaneous and therefore can happen.
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ROadmanKys
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(Original post by charco)
If you read my earlier post I said "You have to compare the half equation given with water with the two cobalt half equations, using E(cell) = E(red) - E(ox) . A spontaneous reaction will have E(cell) positive and greater than +0.3V"

That's how you determine whether or not a reaction is thermodynamically spontaneous (using electrode potentials)

The hydrogen half equation has the electrode potential = 0.0V
The iron(II)|Fe half equation has an electrode potential = -0.44

Hence, for the process:

Fe2+ + H2 --> Fe + 2H+

Using E(cell) = E(red) - E(ox)

E(cell) = -0.44 - 0 = -0.44V

This is negative and therefore the reaction will not happen, it is not spontaneous.

However for the reaction:

2Fe3+ + H2 --> 2Fe2+ + 2H+

E(cell) = +0.77 - 0.0 = +0.77

This reaction is spontaneous and therefore can happen.
in the example you provided with fe2+, why doesnt it oxidise since its E is more negative than water, thus it would be fe -> fe2+ + 2e- ?
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ROadmanKys
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(Original post by charco)
If you read my earlier post I said "You have to compare the half equation given with water with the two cobalt half equations, using E(cell) = E(red) - E(ox) . A spontaneous reaction will have E(cell) positive and greater than +0.3V"

That's how you determine whether or not a reaction is thermodynamically spontaneous (using electrode potentials)

The hydrogen half equation has the electrode potential = 0.0V
The iron(II)|Fe half equation has an electrode potential = -0.44

Hence, for the process:

Fe2+ + H2 --> Fe + 2H+

Using E(cell) = E(red) - E(ox)

E(cell) = -0.44 - 0 = -0.44V

This is negative and therefore the reaction will not happen, it is not spontaneous.

However for the reaction:

2Fe3+ + H2 --> 2Fe2+ + 2H+

E(cell) = +0.77 - 0.0 = +0.77

This reaction is spontaneous and therefore can happen.
Also, with the exam question, first post, why would co not be spontaneous since co2+ + 2e- > co E -0.28 would oxidise, thus in the calculation it would be 1.23 - -0.28 = 1.51 which is higher/more positive than the co3+ since its 0.59, thus more spontanoeus.
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charco
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(Original post by ROadmanKys)
in the example you provided with fe2+, why doesnt it oxidise since its E is more negative than water, thus it would be fe -> fe2+ + 2e- ?
It is not more negative than water ...
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