Calculating a standard potential

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Homework Statement



From the standard potential [tex]Tl^+ + e^-[/tex] -----> [tex]Tl[/tex] (solid) [tex]E^o[/tex][tex]= -0.336 V[/tex]

Determine the standard potential of
[tex]Tl_2S[/tex] (solid) + [tex]2e^-[/tex] -----> [tex]2Tl[/tex] (solid) + [tex]S^2^-[/tex]

Given that the [tex]K_sp[/tex] for [tex]Tl_2S[/tex] is [tex]1.2x10^-22[/tex]

***I cannot get latex to put the -22 in the exponent. The Ksp is 1.2 * 10 ^-22

Homework Equations



Nernst Equation

[tex]E = E^o -(0.05916/n) log(concentration) [/tex]

The Attempt at a Solution




[tex]E = -0.336 -(0.05916/2) log (1.2x10^-22)[/tex]

[tex]E = +0.312 [/tex]

Im wondering if the [tex]log (1.2x10^-22) [/tex] should be [tex]log (1/1.2x10^-22) [/tex]
because the concentration is the products over the reactants. Can someone point me in the right direction? Thanks
 
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Answers and Replies

  • #2
Borek
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10^{-22} = [tex]10^{-22}[/tex]

Write Kso formula for Tl2S and solve for [Tl+].
 
  • #3
ssb
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10^{-22} = [tex]10^{-22}[/tex]

Write Kso formula for Tl2S and solve for [Tl+].

OK BOREK!!! You have given me some insight. Thank you! Products over reactants. and because the reactants are solid, they will be in the denominator as the number 1. Ok I was able to come up with the following calculation. I would be much appreciated if someone could confirm for me the accuracy. Thanks!

Calculate the standard potential ([tex]E_{total}[/tex]) of

[tex] Tl_2S (solid) + 2 e^- = 2Tl (solid) + S^{2-}[/tex]

given that the [tex]K_{sp}[/tex] of [tex]Tl_2S[/tex] is [tex]1.2x10^{-22}[/tex]

[tex]E_{total} = E_{Tl^+/Tl} + E_{ksp} = -0.336 + (0.05916/2) * log(1.2x10^{-22})[/tex]

[tex]E_{total} = -0.336 + (-0.648) = -0.984 [/tex]

so my answer is -0.98 when using significant figures due to the Ksp being 2 sig figs....right?
Look good?
 
  • #4
Borek
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No. Do what I told you to do.
 
  • #5
ssb
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10^{-22} = [tex]10^{-22}[/tex]

Write Kso formula for Tl2S and solve for [Tl+].

Kso formula

[tex]Tl_2S = 2Tl^+ (aq) + S^- (aq)[/tex]

[tex]1.2x10^{-22} = {[Tl^+]^2[S^-]}/[{Tl_2S}][/tex]

[tex]Tl_2S [/tex] is a solid so we can remove it from the equation giving me

[tex]1.2x10^{-22} = {[Tl^+]^2[S^-]}[/tex]

So this is where I get confused... is the correct way to do it this:

[tex]1.2x10^{-22} = {[x]^2[x]}[/tex]

[tex]x = 4.9 x 10^{-8}[/tex]
 
  • #6
Borek
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Solve for Tl+. Not for some x, you are not trying to find out concentration of saturated solution (which you did wrong BTW - concentrations are different, not identical), but concentration of Tl+ as a function of Kso and [S2-].
 
  • #7
ssb
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Solve for Tl+. Not for some x, you are not trying to find out concentration of saturated solution (which you did wrong BTW - concentrations are different, not identical), but concentration of Tl+ as a function of Kso and [S2-].

Borek thank you so much for your help on this problem and that other problem I posted. Borek im going to level with you... I am very confused. I thought I knew what a Kso was (something to do with solubility) but I don't know much more than that. In post # 3 (my first reply) I did something similar to what a TA did on a similar problem and I was following his steps. Im guessing that the problem is where i am taking the log but I really am not positive.

Can you give me a little bit more of a bump in the right direction please? Thanks.
 
  • #8
Borek
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Looks like you have problems with simple algebra.

[tex]K_{so} = [Tl^+]^2[S^{2-}][/tex]

Solving for [Tl+]:

[tex][Tl^+]=\sqrt \frac {K_{so}} {[S^{2-}]}[/tex]

Now put [Tl+] concentration into the original equation. As the question asks about standard potential, sulfide activity is 1.
 
  • #9
ssb
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Looks like you have problems with simple algebra.

[tex]K_{so} = [Tl^+]^2[S^{2-}][/tex]

Solving for [Tl+]:

[tex][Tl^+]=\sqrt \frac {K_{so}} {[S^{2-}]}[/tex]

Now put [Tl+] concentration into the original equation. As the question asks about standard potential, sulfide activity is 1.
Ok ok ok.....

[tex]
E_{total} = E_{Tl^+/Tl} + E_{ksp} = -0.336 + (0.05916/2) * log(\sqrt {1.2x10^{-22}})
[/tex] ?????

This would give
[tex]E_{total} = -0.66 [/tex] ????
 
  • #10
Borek
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Electrode reaction is still Tl+ + e- -> Tl...
 
  • #11
ssb
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Electrode reaction is still Tl+ + e- -> Tl...

omg did i do that honestly


[tex] E_{total} = E_{Tl^+/Tl} + E_{ksp} = -0.336 + (0.05916) * log(\sqrt {1.2x10^{-22}})[/tex]

= -0.984

Look good now?
 
  • #12
Borek
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Better, but still wrong. What sign in Nernst equation if it contains concentration of oxidized form?
 
  • #13
ssb
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Better, but still wrong. What sign in Nernst equation if it contains concentration of oxidized form?

It would make it negative wouldnt it?

[tex] E_{total} = E_{Tl^+/Tl} + E_{ksp} = -0.336 + (\frac {.05916}{-1}) * log(\sqrt {1.2x10^{-22}}) [/tex]

= 0.312 volts

with sig figs making the answer 0.31 volts

This has to be correct this time. (btw thanks you have been unknowingly teaching me latex as well!)
 
  • #14
Borek
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Oops, sorry. You have been juggling signs and log argument and at about 1 a.m. you've lost me. -0.984 was OK. Still, your original Nernst equation

[tex]E = E^o -(0.05916/n) log(concentration) [/tex]

is incorrect, as long as it doesn't state concentration of what.

[tex]E = E_0 + \frac {RT} {nF} ln {\frac {[Ox]} {[Red]}}[/tex]

if reduced form is solid, it simplifies to

[tex]E = E_0 + \frac {RT} {nF} ln {[Ox]}[/tex]

or

[tex]E = E_0 + \frac {0.05916} {n} log {[Ox]}[/tex]
 

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