Standard Reduction Potential Notation?

[tncekm]

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Okay, so I saw some info with reduction potentials compared, but the notation didn't make sense to me. Maybe it was part of an equation I'm not familiar with?

Anyway, any help is much appreciated.

E&#8304;(Zn²&#8314;/Zn) <--- WTF is this?

 

[Vihsadas]

I can't really see what exactly those superscripts are, but I think its just:

Eo(Zn2+/Zn)

Which probably means, 'The standard reduction potential to convert Zn2+ to Zn'.

 

[Kaustikos]

I can't really see what exactly those superscripts are, but I think its just:

Eo(Zn2+/Zn)

Which probably means, 'The standard reduction potential to convert Zn2+ to Zn'.

That's what I was going to say. I dont' know why they'd add the standard superscripts to the elements, which would mean that it equals 0, right? The standard something of an element in its standard state is always 0....

 

[Zerconia2921]

Okay, so I saw some info with reduction potentials compared, but the notation didn't make sense to me. Maybe it was part of an equation I'm not familiar with?

Anyway, any help is much appreciated.

E&#8304;(Zn²&#8314;/Zn) <--- WTF is this?

The standard reduction potentials is E = reduction + oxdation

For a reaction to occur there is greater potential for reduction than oxidation. Which means the more positive the E the greater chance for reduction to occur. I hope i made sense.

 

[tncekm]

I can't really see what exactly those superscripts are, but I think its just:

Eo(Zn2+/Zn)

Which probably means, 'The standard reduction potential to convert Zn2+ to Zn'.

That's what I was going to say. I dont' know why they'd add the standard superscripts to the elements, which would mean that it equals 0, right? The standard something of an element in its standard state is always 0....

Thanks, and yeah Vihsadas, you got the superscripts correct (I guess the characters my mac defined didn't want to show on your comp?). I've just never seen notation like that, and it kind of threw me. I'm used to seeing X+ + e- ----> X2; E=+Z V, etc.

 

[tncekm]

The standard reduction potentials is E = reduction + oxdation

For a reaction to occur there is greater potential for reduction than oxidation. Which means the more positive the E the greater chance for reduction to occur. I hope i made sense.

Thanks, but I'm not quite following.

I thought for a voltaic reaction to occur there just needed to be a spontaneous flow of electrons from one molecule to an atom (a reductant reduces something, and is oxidized spontaneously).

Could you explain the E = reduction + oxidation? I thought E is just the "electromotive force", which has to do with electron movement where positive electron movement is spontaneous and negative is forced (like an electrolytic cell).

 

[Zerconia2921]

Thanks, but I'm not quite following.

I thought for a voltaic reaction to occur there just needed to be a spontaneous flow of electrons from one molecule to an atom (a reductant reduces something, and is oxidized spontaneously).

Could you explain the E = reduction + oxidation? I thought E is just the "electromotive force", which has to do with electron movement where positive electron movement is spontaneous and negative is forced (like an electrolytic cell).

E is the electromotive force or you can also have EMF which is the same.
EMF= reduction + oxidation this equation is the difference in potentials between the two half cells.

For a galvanic cell the EMF is positive whereas for an electrolytic cell its negative.

For example:

Eq. 1) Co3+ + e- --> Co2+ E = +1.82 Volts
Eq. 2) Na+ + e- --> Na(s) E = -2.71 Volts

The larger the potential the more likly reduction occurs. So looking at the two equations which one has the larger value?....

Eq.1) correct its value of 1.82 Volts is > -2.71 Volts. So the total reaction will look like this:

Co3+ + e- + Na(s) ---> Na+ + e- + Co2+

Since we reversed equation 2 then the E value will change to a positive 2.71 volts.

So your EMF will be 1.82 + 2.71 volts = 4.53 volts. Since our EMF is positive it shows to be a spontaneous reaction. I hope I answered your question.

 

[BerkReviewTeach]

Okay, so I saw some info with reduction potentials compared, but the notation didn't make sense to me. Maybe it was part of an equation I'm not familiar with?

Anyway, any help is much appreciated.

E&#730;(Zn&#178;/Zn) <--- WTF is this?

It's outdated shorthand for a half-reaction, exactly as Vihsadas and Kaustikos pointed out. It goes along with cell shorthand. For an electrochemical cell, they often use shorthand such as:

Zn|0.10 M Zn2+||0.10 M Cu2+|Cu

I assume such a thing is present in your review materials because the author wants to expose you to as many different ways to present the material as possible.

The standard reduction potentials is E = reduction + oxdation

You really need to be careful in terms of terminology, because the MCAT preys on careless errors. The standard reduction potential is for a reduction HALF-reaction. It is measured as a voltage and can be referred to as an emf in terms of voltage. But it is only a half-reaction.

What you have typed is a standard reaction potential. Ereaction = Ereduction + Eoxidation is for a full redox reaction. It's a subtle difference, but one that you can be sure an incorrect answer is based on.

 

[TawMus]

It's outdated shorthand for a half-reaction, exactly as Vihsadas and Kaustikos pointed out. It goes along with cell shorthand. For an electrochemical cell, they often use shorthand such as:

Zn|0.10 M Zn2+||0.10 M Cu2+|Cu

I assume such a thing is present in your review materials because the author wants to expose you to as many different ways to present the material as possible.



You really need to be careful in terms of terminology, because the MCAT preys on careless errors. The standard reduction potential is for a reduction HALF-reaction. It is measured as a voltage and can be referred to as an emf in terms of voltage. But it is only a half-reaction.

What you have typed is a standard reaction potential. Ereaction = Ereduction + Eoxidation is for a full redox reaction. It's a subtle difference, but one that you can be sure an incorrect answer is based on.

What is the equation for the standard reduction potential then???