[Renaud]

Hi everyone ,

I have a question regarding the effect of adding 26 mL NaOH 0.09 mol/L to the sulfur compartment of the following battery:
Hg 2+ (0.02mol/L) | Hg | SO42- (0.2 mol/L) | S2O62-

We are asked to calculate the new potential after this addition.

We're at pH 2.

The half reactions are as follows:

• Hg 2+ + 2e- <--> Hg (E°=0.851)
  2SO4 2- + 4H+ + 2e- <--> S2O62- + 2H2O (E°=-0,2)

The reaction of the battery: Hg 2+ + S2O62- + 2H2O --> Hg + 2SO4 2- + 4H+

Will adding NaOH just change the sulphur concentration, or will the basic effect of NaOH affect the potential of the battery?

Number of moles of NaOH = 2.34 E-3

Logically, the concentration of SO42- should change to 0.2/(1+26 E-3) = 0.195 mol/L, from there I simply modify the concentration in the Nernst equation?

Thanks in advance for your help

[mjc123]

Do you know the volume of solution is 1L? You have to know the volume to answer the question.
What is the effect on the pH of adding the NaOH? How does that affect the potential?

[Renaud]

I don't have any volume but let's say it is 1L.
Concentration of H3O+ is 0,01, we know this by knowing the pH value.

Before adding the NaOH :

Moles of H+ in the solution = 0,01

After adding the NaOH :

Moles of NaOH added = 2,34 E-3

Moles of H+ in the solution after reacting with NaOH, Moles of H+ : 7,66 E-3 (0,01 - 2,34 E-3)

Nernst equation  attached (0,05 is the concentration of S2O62-

New potential : 1,254 V

Is that correct ? 

[mjc123]

I don't have any volume but let's say it is 1L.

You must have the volume to be able to answer the question. If they asked the question without giving you the volume, they are idiots.

0,05 is the concentration of S2O62-

You kept quiet about that! Anything else you didn't tell us - like the volume?

New potential : 1,254 V

I agree. Can you work out what it was before adding the NaOH, so you can see the effect of the addition?
You might also like to try correcting for the change of volume - see what effect that has, compared to the effect of changing pH.

[Renaud]

The S2O62- and the volume were not indicated in the exercise, but that's a mistake.
The potential increases from 1.24 to 1.125 V thanks to the diminution of the H+ ions

H++OH- -> H2O
0.01 moles of H+ in the begining, and we add 0.00234 moles of NaOH so,

0.01 - 0.00234 = 0.00766 moles of H+ remaining in the solution

Just replace 0,01 by 0.00766 in the Nernst equation