14.33 (b)


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Andy Liao 1B
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Joined: Thu Jul 13, 2017 3:00 am

14.33 (b)

Postby Andy Liao 1B » Sat Feb 17, 2018 11:53 pm

Problem:
14.33 (a) The standard Gibbs free energy of formation of Tl3+(aq) is +215 kJ.mol-1 at 25°C. Calculate the standard potential of the Tl3+/Tl couple. (b) Will Tl+ disproportionate in aqueous solution?

I don't understand why in the solutions manual the equation of interest is 3Tl+(aq) -> 2Tl(s) + Tl3+(aq). Can someone please explain?

sahiltelang-Discussion 1J
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Re: 14.33 (b)

Postby sahiltelang-Discussion 1J » Sun Feb 18, 2018 9:57 am

They derive this equation by using Appendix 2B and coupling the half rxns for the oxidation and reduction of T1+.

Dang Lam
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Joined: Thu Jul 27, 2017 3:01 am

Re: 14.33 (b)

Postby Dang Lam » Mon Feb 19, 2018 12:15 am

for part a, why did we have to flip the sign of delta G to negative instead of keeping it positive like given? Also how did they come up with the equation for part a?

Farah Ahmad 2A
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Re: 14.33 (b)

Postby Farah Ahmad 2A » Mon Feb 19, 2018 6:46 pm

You flip the sign of deltaG because the deltaG given is for the formation of Tl3+. The question asks for the standard potential of the Tl3+ /Tl couple, which when written like that indicates that Tl3+ is the reactant and Tl is the product, making it the flipped version of the formation of Tl3+. That is why we have to make +215 kJ/mol into -215kJ/mol.

Dang Lam
Posts: 55
Joined: Thu Jul 27, 2017 3:01 am

Re: 14.33 (b)

Postby Dang Lam » Mon Feb 19, 2018 10:11 pm

Farah Ahmad 2A wrote:You flip the sign of deltaG because the deltaG given is for the formation of Tl3+. The question asks for the standard potential of the Tl3+ /Tl couple, which when written like that indicates that Tl3+ is the reactant and Tl is the product, making it the flipped version of the formation of Tl3+. That is why we have to make +215 kJ/mol into -215kJ/mol.

Got it! thank you


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