Standard electrode potential calculation

This is one I am repeatedly screwing up.

Say we have two half reactions:

V³⁺ + e^- --> V²⁺ E of half cell = -0.26V
VO²⁺ + 2H⁺ + e^- --> V³⁺ + H₂O E of half cell = +0.34V

Obviously the site of oxidation would be the first half cell as its E is more negative. So the V²⁺ would be oxidised to form V³⁺ ions and the resulting electrons are used to reduce VO²⁺ - essentially the first half cell would go backwards.

Ecell is calculated using E(most positive) - E(least positive). However, because the first half cell occurs backwards, do we reverse its E value to +0.26V, or leave it alone?

That confuses me because surely if it occurs backwards the voltage must be the negative of what it would be in the forward reaction.

Also if we have to double quantities of one half cell for the overall reaction to balance, do we double the E of that half cell in calculation?

Cheers

(edited 9 years ago)

Reply 1

Vadevalor

3

1) leave it alone
2) yes double it makin sure resulting ionic eqn is rxn u want

Reply 2

spleenharvester

OP

18

Original post by Vadevalor

1) leave it alone
2) yes double it makin sure resulting ionic eqn is rxn u want

Awesome, thankyou. :smile:

Reply 3

Vadevalor

3

Btw when you form resulting equation, invert the first half equation so the electron is on the right. *Dont invert sign but invert equation*

Reply 4

KanKan

11

Original post by spleenharvester

This is one I am repeatedly screwing up.

Say we have two half reactions:

V³⁺ + e^- --> V²⁺ E of half cell = -0.26V
VO²⁺ + 2H⁺ + e^- --> V³⁺ + H₂O E of half cell = +0.34V

Obviously the site of oxidation would be the first half cell as its E is more negative. So the V²⁺ would be oxidised to form V³⁺ ions and the resulting electrons are used to reduce VO²⁺ - essentially the first half cell would go backwards.

Ecell is calculated using E(most positive) - E(least positive). However, because the first half cell occurs backwards, do we reverse its E value to +0.26V, or leave it alone?

That confuses me because surely if it occurs backwards the voltage must be the negative of what it would be in the forward reaction.

Also if we have to double quantities of one half cell for the overall reaction to balance, do we double the E of that half cell in calculation?

Cheers

You leave it alone because the calculation has already taken that into account for you.

And you do not double the E of that half cell! Just because the overall reaction requires the stoichiometry to be doubled, does not mean the potential difference (voltage) is doubled. It stays the same! :smile:

Reply 5

spleenharvester

OP

18

Just need to clarify, do we or don't we double the potential when doubling the equation?

Cause surely, for example, Ca ---> Ca2+ and 2e- doubled to 2Ca ---> 2Ca2+ and 4e-, in the first case, there are two electrons on the right and none on the left, whereas in the second case there are four electrons on the right and none on the left, surely it would double as well?

Cheers :smile:

Reply 6

Borek

4

Original post by spleenharvester

Just need to clarify, do we or don't we double the potential when doubling the equation?