Calculating a standard potential

[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...

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

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

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