I need a chemistry geek

[jessiel8]

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#24 on aamc 7
Electrochemistry question.

When you are trying to find Ecell, i understand that you add Eox + Ered potentials for the two half reactions.

However, when the question gives you the E values for the two half-reactions, do you have to reverse the sign of the E value for the oxidation half reaction? (in the answer key they dont, they just add the two numbers....i thought you always had to!?!)
Please help, thanks

 

[buffdoc]

Just look at the overall reaction and compare it to the half cell reactions. If you have to reverse one of the half cell reactions to make it fit into the overall reaction, then you must change the sign of the half cell electrode potential. Then add the two half cell potentials.

 

[MasterMD]

they are both in the correct direction to do just straight addition 😉

 

[grapeflavorsoda]

#24 on aamc 7
Electrochemistry question.

When you are trying to find Ecell, i understand that you add Eox + Ered potentials for the two half reactions.

However, when the question gives you the E values for the two half-reactions, do you have to reverse the sign of the E value for the oxidation half reaction? (in the answer key they dont, they just add the two numbers....i thought you always had to!?!)
Please help, thanks

generally, when the question stem does not give any explicit clues to whether the E cell is galvanic or electrolytic, see the reaction equation and determine which is oxidized and reduced. After determining that, use the reduction potential of the reduced substance as given, but use negative value of the reduction potential for the oxidized, to get the sum which gives you the cell potential(asuming that it is in the standard state).

But if the question stem gives you the reaction equation and relevant half reaction potetials but not the "reduction" potential, you dont have to switch the sign for oxidation half reaction since doing so will give you the potential for the opposite reaction.

you have to remember the definition of the reduction potential. It is a "relative potential" of the "reduction reaction" in comparison to some arbitrary but fixed half reaction(the standard hydrogen reduction rxn). The (standard) reduction potential is nothing more than relative free energy change per electron transferred to some ion or compounds. Since the (Gibbs) free energy is a state function, when you reverse the path of a particular reaction the magnitude of the free reaction does not change but the sign reverses. This is why one has to switch the sign of the "reduction potential" given to calculate the potential of the relevant "oxidation reaction".
And also, if the value of the oxidation potential of a particular half reaction is given, you dont have to switch the sign of the oxidation potential to get the oxidation potential of relevant half reaction since doing so will give you value for the reduction potential.

Since the question you are asking seems to give the potential but not standard potential, you have to treat is a bit differently than question that gives you only the reduction potential.

See whether one of the half reaction potential given is oxidation (based on your reference to the aamc solution, one of them has to be oxidation potential and the other being reduction potential).

 

[rgerwin]

This question confused me, too, b/c I could swear there were 2 moles of one species to 1 mol of the other, so one of the half-cell voltages need to be doubled, but the answer didn't show that. Any comments?

 

[grapeflavorsoda]

This question confused me, too, b/c I could swear there were 2 moles of one species to 1 mol of the other, so one of the half-cell voltages need to be doubled, but the answer didn't show that. Any comments?

Again, remember the definition correctly.

Potential is the relative change of the free energy per electron.

If if the reaction equation involves 2 moles, 100moles, or whatever of the half reaction species, when you calculate the reaction potential you are calculating free energy per electron and given referential potentials are free energy changes per electron.

You just add the relevant half reaction potentials without doing any calculation on the potential values besides multiplying negative 1 to relevant half reactions.

As a general rule, there is no need to multiply any number that has absolute value not equal to 1, while you are doing calculation using half reactions.

Here is simple proof.

before the proof i will define some notations.

Let e = electron, A(x+) = cation A having postive "x" charge, B(y+) = cation B having positive y charge, and Let A and B be the neutral species of the respective cations.

Assume some redox reaction involving following equation;

yA(x+) + xB -----> yA + xB(y+) and let this reaction have the reaction potential E(rxn).

Notice that the above reaction involves xy (x multiplied by y) electrons.

Further assume the relevant half reaction have the following potentials(i am not using reduction potentials for the sake of the simplicity but same principle applies to the reduction potential based).

A(x+) + xe -----> A ; E = C

B -----> B(y+) + ye ; E = D

In order to calculate the potential energy of the reaction( not the reaction potential which defined as the potential energy of the reaction per electron involved), you need to multiply each potential of half reaction by the total number of the electron involved which is "xy" .

Adding the resulting values will give the potential energy of the redox reaction in question.

yA(x+) + xB -----> yA + xB(y+) ; Change of the free energy = xyC + xyD and

the sum xyC + xyD = xy(C+D).

Since the reaction potential of the above redox reaction is defined as the change of the free energy of the reaction per electron involved, we can get the Reaction potential, E(rxn) = {xy(C+D)}/xy = C+D.

Notice that the value C+D is just sum of the relevant potential for the half reaction.

Thus the general rule saying "there is no need to multiply any number that has absolute value not equal to 1, while you are doing calculation using half reactions" is true.

End of the proof.

And this short proof just reminded me why i decided to quit visiting this site after last year's MCAT. 😛

 

[MasterMD]

This question confused me, too, b/c I could swear there were 2 moles of one species to 1 mol of the other, so one of the half-cell voltages need to be doubled, but the answer didn't show that. Any comments?

😱 never NEVER mutiply potentials by coefficients.... it DOES NOT change with how much you have!!!!! go with whatever they give you except for the +/- which may need to be reversed depending on how the reaction adds up from the half reactions...