Determine the equilibrium constant (Keq) at 25°C for the reaction

Cl2(g) + 2Br– (aq) 2Cl– (aq) + Br2(l)

A. 6.3 × 109

B. 1.5 × 10–10

C. 1.3 × 1041

D. 8.1 × 104

E. 9.8

Also,

Determine the equilibrium constant, Keq, at 25°C for the reaction

2Br– (aq) + I2(s) Br2(l) + 2I– (aq)

A. 5.7 × 10–19

B. 5.7 × 10–55

C. 1.7 × 1054

D. 1.9 × 1018

E. 18.30

Please and thank you so much :)

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## Answers & Comments

## Verified answer

You can use the standard reduction half reactions to find Ecell and then you can relate Ecell to Keq (Normally the half reactions or Ecell would be given in a problem like this). If you figure out the half reactions. You can look them up in standard reduction potential chart. From the half reactions you need to find the Ecell. Then using RTln(Keq) = nF(Ecell) you can solve for Keq.

There is also a different way to solve it by adding half reactions straight up, but that gets a bit confusing when keeping track of "n" or the amount of electrons. In this case however it's easy because there are only 2e- for each half reaction

Also, if you are given a Delta G or a way to solve for delta G for the reaction you can use Delta G = -RTlnKeq and you should get the same answer.

EDIT: I'm actually just sitting around bored so I went through and worked them out

Cl2 + 2e- = 2Cl- E=1.36 V

Br2 + 2e- = 2Br- E= 1.07 V

Ecell = Ered - Eox

Ecell = 1.36 - 1.07 = .29 V

In this case n = 2 (Both half potentials have 2e-) This makes it slightly more easy.

ln Keq = [nF(Ecell)] / RT

Keq = e^[(Ecell *n)/0.025693V]

Keq = e^[(0.29 * 2)/0.025693V]

Keq = 6.4 * 10^9 (Answer A. I probably used a slightly different half potential or rounded my constants differently)

Double checking, This makes sense. A more positive reduction potential means that the half reaction favors the right hand side (it wants the electrons) Therefore, it makes sense that the Keq > 1 because the Cl2 reduction potential is higher than the Br2 one at standard conditions (meaning that Cl2 wants electrons more than Br2 does) Keep in mind that as soon as you put this together at standard concentrations/pressures it will begin to react and concentrations/pressures change and this calculation also changes as a result.

You solve the 2nd one the exact same way but you use the I2 to iodide reduction potential instead.

I2 + 2e- = 2I- E=0.54 Already we can predict that the Keq will be smaller than 1 because the E for this half reaction is less positive meaning I2 wants electrons less than Br2 and the reverse reaction will be favored.

In this case Ecell = Ered + Eox = 0.54 - 1.07 = -0.53 (Notice the negative Ecell value)

Using the same formula as above

We solve for Keq and get 1.21 * 10^-18 I'm not sure if i used different numbers for reduction potentials but I'm not getting any of the answers you have. I very easily could have made a mistake. However, this answer makes sense to me.

Hope this helps you. Cheers

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