[Shaare]

Hi, struggling to get my head around this question.

It says the E-nought value for Cr207- + 14H+ + 6e- ----> 2Cr3+ + 7H20 is +1.33V.
The question asks me to use the E-nought value of Chlorine being reduced to Chloride ions (+1.36V) to show that the Cr2072- will not oxidise the Cl- ions.

Since the question is asking about chloride ions I have reversed the equation and used the E-nought value as -1.36V. Using the standard convention of putting the most negative electrode on the left I have proceeded to do E-nought(RHS) - E-nought(LHS) which is 1.33-(-1.36) = +2.69V. This shows the reaction to be feasible.

However in the book it says the E-nought(cell) is -0.03V. I can't see where I have gone wrong with my method as it hasn't failed me before.

Any help would be greatly appreciated. Thanks!

[gearoid94]

Whenever I do these I don't use that equation I just get the 2 values and add them together which will give you the right answer.

[Shaare]

(Original post by gearoid94)
Whenever I do these I don't use that equation I just get the 2 values and add them together which will give you the right answer.

Why do you add them?

[gearoid94]

That's how we were tought to do them, we don't use that equation. I think the reason it doesn't work is because that equation is for finding the voltage of a cell whereas this is just a reaction.

[Shaare]

Ah, I see. Thanks.

Still a little confused, can anyone else shed some light?

[charco]

(Original post by Shaare)
Ah, I see. Thanks.

Still a little confused, can anyone else shed some light?

OK - here's the method I propose:

NEVER change the values of the standard reduction potentials, even if you reverse the equation.

If you wish to know whether a reaction can procede identify the species that would get reduced and the species that would get oxidised and then apply the formula E = E(reduced species) - E(oxidised species)

You can abbreviate the equation as E = E(RED-OX) to help memory.

Now let's apply it to your problem:

If yo wish to use dichromate ions to oxidise chloride ions, the dichromate is the species that gets reduced and the chloride gets oxidised.

E = E(RED-OX)

E = 1.33 - 1.36 = -0.03V

As the answer is negative the reaction is not spontaneous, i.e. cannot happen under standard conditions.

As a rule of thumb values for E(cell) between 0 and +0.3 are spontaneous, but arrive at equilibrium (under standard conditions).

[Shaare]

(Original post by charco)
OK - here's the method I propose:

NEVER change the values of the standard reduction potentials, even if you reverse the equation.

If you wish to know whether a reaction can procede identify the species that would get reduced and the species that would get oxidised and then apply the formula E = E(reduced species) - E(oxidised species)

You can abbreviate the equation as E = E(RED-OX) to help memory.

Now let's apply it to your problem:

If yo wish to use dichromate ions to oxidise chloride ions, the dichromate is the species that gets reduced and the chloride gets oxidised.

E = E(RED-OX)

E = 1.33 - 1.36 = -0.03V

As the answer is negative the reaction is not spontaneous, i.e. cannot happen under standard conditions.

As a rule of thumb values for E(cell) between 0 and +0.3 are spontaneous, but arrive at equilibrium (under standard conditions).

Does this rule only apply for the feasibility of reactions?

For example, how would you work out the E(cell) for Zn|Zn2+ || Pb2+|Pb?

Given that Zn2+ + 2e- ---> Zn [E(cell) = -0.76V]
Pb2+ + 2e- ---> Pb [E(cell) = -0.13V]

Using your method, E(red) is -0.76V and E(ox) is -0.13V.

-0.76-(-0.13) = -0.63V.

However the answer in the book is +0.63V and the answer I get actually corresponds to part b of the question which asks for the E(cell) for Pb|Pb2+ || Zn2+||Zn.

[charco]

(Original post by Shaare)
Does this rule only apply for the feasibility of reactions?

For example, how would you work out the E(cell) for Zn|Zn2+ || Pb2+|Pb?

Given that Zn2+ + 2e- ---> Zn [E(cell) = -0.76V]
Pb2+ + 2e- ---> Pb [E(cell) = -0.13V]

Using your method, E(red) is -0.76V and E(ox) is -0.13V.

-0.76-(-0.13) = -0.63V.

However the answer in the book is +0.63V and the answer I get actually corresponds to part b of the question which asks for the E(cell) for Pb|Pb2+ || Zn2+||Zn.

In reality, it is not possible to have a cell emf that is a negative value. It is calculated as the difference between two values (the electrode potentials under the conditions given).

The negative sign just shows you the direction of electron flow around the external circuit.

In the example that you have given, you show that the reaction between Pb and Zn²⁺ is NOT feasible, so how can you generate a voltage from the cell?

What you really mean to say is that the reduced species (Pb²⁺) receives electrons from the oxidised species (Zn) and the cell potential is +0.63V with electrons flowing from the zinc half cell to the lead half cell.

[Shaare]

(Original post by charco)
In reality, it is not possible to have a cell emf that is a negative value. It is calculated as the difference between two values (the electrode potentials under the conditions given).

The negative sign just shows you the direction of electron flow around the external circuit.

In the example that you have given, you show that the reaction between Pb and Zn²⁺ is NOT feasible, so how can you generate a voltage from the cell?

What you really mean to say is that the reduced species (Pb²⁺) receives electrons from the oxidised species (Zn) and the cell potential is +0.63V with electrons flowing from the zinc half cell to the lead half cell.

Thank you, I think I get what you mean. Sorry for being 4 weeks late .