What Is the Standard Potential for Tl2S Reduction?

[ssb]

Homework Statement

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

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

Given that the K_sp for Tl_2S is 1.2x10^-22

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

Homework Equations

Nernst Equation

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

The Attempt at a Solution

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

E = +0.312

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

 

[Borek]

10^{-22} = 10^{-22}

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

 

[ssb]

10^{-22} = 10^{-22}

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 (E_{total}) of

Tl_2S (solid) + 2 e^- = 2Tl (solid) + S^{2-}

given that the K_{sp} of Tl_2S is 1.2x10^{-22}

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

E_{total} = -0.336 + (-0.648) = -0.984

so my answer is -0.98 when using significant figures due to the Ksp being 2 sig figs...right?
Look good?

 

[Borek]

No. Do what I told you to do.

 

[ssb]

10^{-22} = 10^{-22}

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

Kso formula

Tl_2S = 2Tl^+ (aq) + S^- (aq)

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

Tl_2S is a solid so we can remove it from the equation giving me

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

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

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

x = 4.9 x 10^{-8}

 

[Borek]

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 [S^2-].

 

[ssb]

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 [S^2-].

Borek thank you so much for your help on this problem and that other problem I posted. Borek I am 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. I am 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.

 

[Borek]

Looks like you have problems with simple algebra.

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

Solving for [Tl⁺]:

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

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

 

[ssb]

Looks like you have problems with simple algebra.

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

Solving for [Tl⁺]:

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

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

Ok ok ok...

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

This would give
E_{total} = -0.66 ?

 

[Borek]

Electrode reaction is still Tl⁺ + e^- -> Tl...

 

[ssb]

Electrode reaction is still Tl⁺ + e^- -> Tl...

omg did i do that honestly


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

= -0.984

Look good now?

 

[Borek]

Better, but still wrong. What sign in Nernst equation if it contains concentration of oxidized form?

 

[ssb]

Better, but still wrong. What sign in Nernst equation if it contains concentration of oxidized form?

It would make it negative wouldn't it?

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

= 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!)

 

[Borek]

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

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

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

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

if reduced form is solid, it simplifies to

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

or

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