6.57


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Joseph Saba
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Joined: Thu Jul 11, 2019 12:16 am

6.57

Postby Joseph Saba » Tue Feb 25, 2020 10:19 pm

6.57 Use the data in Appendix 2B and the fact that, for the half- reaction F2(g) + 2 H+(aq) + 2 e- ---> 2 HF(aq), E(knot)= +3.03 V, to calculate the value of Ka for HF.
How would I do this problem?

Benjamin Feng 1B
Posts: 102
Joined: Sat Sep 07, 2019 12:19 am

Re: 6.57

Postby Benjamin Feng 1B » Wed Feb 26, 2020 1:02 am

You can relate E to Gnot using Gnot = -nFE. From there, relate Gnot to K using Gnot = -RT lnK. In this case, the reaction given makes K equal to Ka

705121606
Posts: 68
Joined: Wed Sep 18, 2019 12:17 am

Re: 6.57

Postby 705121606 » Sat Feb 29, 2020 7:13 pm

Benjamin Feng 1B wrote:You can relate E to Gnot using Gnot = -nFE. From there, relate Gnot to K using Gnot = -RT lnK. In this case, the reaction given makes K equal to Ka


I solved up until K but didn't know how to get to Ka. In the solutions manual they took the square root of K to solve for Ka. Do you know why this is the case?

Emma Popescu 1L
Posts: 105
Joined: Wed Sep 11, 2019 12:16 am

Re: 6.57

Postby Emma Popescu 1L » Sat Feb 29, 2020 10:23 pm

Using the appendix you find that F2+ 2e- -> 2F- gives a value of Enot of +2.87V. Then by using the equation Enot (cathode)- Enot (anode), you can find Enot cell. Then using the equation InK=nFEnot/RT you can find K. To find Ka, just find the square root of K.


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