HW 6.57


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Hannah Lee 2F
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Joined: Thu Jul 11, 2019 12:15 am

HW 6.57

Postby Hannah Lee 2F » Mon Mar 02, 2020 8:50 pm

Use the data in Appendix 2B and the fact that, for the half- reaction F2 + 2H+ + 2e- --> 2HF with E = +3.03 V, to calculate the value of Ka for HF.

Why do you have to take the square root of K at the end to find Ka? I thought the acid disassociation constant was basically an equilibrium constant, just for an acid disassociation reaction, and assumed they would be the same.

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

Re: HW 6.57

Postby Benjamin Feng 1B » Mon Mar 02, 2020 10:54 pm

By definition, the Ka for the reaction would be in the form [H+][F-]/[HF]. When you end up adding the F- half reaction and calculate K, the equation for K will look like [H+]^2, as in the half reaction below, there is a coefficient of 2 in front of H+. To solve for Ka, you can just square root.

Robert Tran 1B
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Joined: Thu Jul 11, 2019 12:15 am

Re: HW 6.57

Postby Robert Tran 1B » Mon Mar 02, 2020 11:32 pm

When you add the half reactions together, you get a coefficient of 2 in front of every reactant and product. In the dissociation reaction, the stoichiometric coefficients are 1. Thus, in order for the Ka to correspond to a stoichiometric coefficient of 1, we must square root the K value.

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

Re: HW 6.57

Postby DHavo_1E » Thu Mar 12, 2020 9:10 am

Hannah Lee 2F wrote:Use the data in Appendix 2B and the fact that, for the half- reaction F2 + 2H+ + 2e- --> 2HF with E = +3.03 V, to calculate the value of Ka for HF.

Why do you have to take the square root of K at the end to find Ka? I thought the acid disassociation constant was basically an equilibrium constant, just for an acid disassociation reaction, and assumed they would be the same.



Hello,

Can I also ask how you determined which was the anode and which was the cathode for this problem? Thank you!

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

Re: HW 6.57

Postby Hannah Lee 2F » Thu Mar 12, 2020 5:33 pm

By definition, disassociation means that the chemical equation will take the form AB --> A + B. So in this case, we want the net equation for the disassociation of the acid HF to be HF --> H+ + F-, and we need to adjust our anode and cathode half-reactions accordingly.

In order for HF to be on the reactant side, we need to flip the reduction half-rxn (F2 + 2H+....) given to us, so we know it is the anode. To cancel out the F2 and add a F- to the product side, we can look at the appendix to find the reduction reaction: F2 + 2e- --> 2F-.

Since the reduction potentials of the anode was already provided for us, and we can look up the reduction potential of the cathode in the appendix, we find out that the overall cell potential is: Ecell = Ecath - Eanode = 2.87 - 3.03 V = -0.16. Since the standard cell potential is negative, we can imply that the rxn will be non-spontaneous under standard conditions, and based on the Nernst equation, we know that [R] > [P].

You can solve the rest of the problem by calculating for K using the Nernst equation under equilibrium conditions.

aishwarya_atmakuri
Posts: 101
Joined: Sat Jul 20, 2019 12:15 am

Re: HW 6.57

Postby aishwarya_atmakuri » Thu Mar 12, 2020 5:58 pm

If you write the Ka out, it is [H+][A-] and in order to get [H+], you would need to take the square root of the value.

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

Re: HW 6.57

Postby DHavo_1E » Fri Mar 13, 2020 8:12 pm

Hannah Lee 2F wrote:By definition, disassociation means that the chemical equation will take the form AB --> A + B. So in this case, we want the net equation for the disassociation of the acid HF to be HF --> H+ + F-, and we need to adjust our anode and cathode half-reactions accordingly.

In order for HF to be on the reactant side, we need to flip the reduction half-rxn (F2 + 2H+....) given to us, so we know it is the anode. To cancel out the F2 and add a F- to the product side, we can look at the appendix to find the reduction reaction: F2 + 2e- --> 2F-.

Since the reduction potentials of the anode was already provided for us, and we can look up the reduction potential of the cathode in the appendix, we find out that the overall cell potential is: Ecell = Ecath - Eanode = 2.87 - 3.03 V = -0.16. Since the standard cell potential is negative, we can imply that the rxn will be non-spontaneous under standard conditions, and based on the Nernst equation, we know that [R] > [P].

You can solve the rest of the problem by calculating for K using the Nernst equation under equilibrium conditions.



Hello,

I am a bit confused about your ending statement regarding how [R]>[P] based on the Nernst equation. When you say that standard cell potential is negative, would that not mean that in the Nernst equation: Ecell = Enaughtcell - RT/nF lnQ, Ecell would be positive since Enaughtcell is negative, and lnQ is also negative (so a negative minus a negative means a negative added to a positive)? Thank you for clearing up about the anode and cathode though!


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