# Calculating a standard potential

1. Jul 28, 2008

### ssb

1. The problem statement, all variables and given/known data

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
2. Relevant equations

Nernst Equation

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

3. 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

Last edited: Jul 28, 2008
2. Jul 29, 2008

### Staff: Mentor

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

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

3. Jul 29, 2008

### ssb

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?

4. Jul 29, 2008

### Staff: Mentor

No. Do what I told you to do.

5. Jul 29, 2008

### ssb

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:

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

$$x = 4.9 x 10^{-8}$$

6. Jul 29, 2008

### Staff: Mentor

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. Jul 29, 2008

### ssb

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. Jul 29, 2008

### Staff: Mentor

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.

9. Jul 29, 2008

### ssb

Ok ok ok.....

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

This would give
$$E_{total} = -0.66$$ ????

10. Jul 29, 2008

### Staff: Mentor

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

11. Jul 29, 2008

### ssb

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?

12. Jul 29, 2008

### Staff: Mentor

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

13. Jul 29, 2008

### ssb

It would make it negative wouldnt 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!)

14. Jul 30, 2008

### Staff: Mentor

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

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

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

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