[LHM]

1. Which statement is not true about the electrolysis of a 1 M solution of KI to which phenolphthalein has been added?

 A) Potassium metal is formed.
 B) A yellow color appears at the anode.
 C) A pink color appears at the cathode.
 D) A gas is produced at the cathode.

The answer is A, but I really don't know how you get the answer, and I've never really encountered adding phenolphthalein to an electrolysis, since I'm bad at electrolysis in the first place  

So if someone could explain the reasoning I would really appreciate it?

2. Ni (s) + Cu²⁺ (aq)  Ni²⁺ (aq) + Cu (s)
The voltaic cell based on this reaction has a voltage of 0.59 V under standard conditions. Which of these changes will produce a higher voltage?
           I. increasing [Cu²⁺]
           II. increasing the size of the Ni(s) electrode

 A) I only
 B) II only
 C) Both I and II
 D) Neither I nor II

The answer's A. I can see why II won't produce a higher voltage, but why does I. work?


Thanks.

[JadenErius]

As to your first question it is impossible for such a low concentration of KI to produce potassium metal. the answer seems valid. I think the pink color could be due to the release of hydrogen ions thereby a larger amount of hydroxide ions would be present (higher pH) which would turn phenolphthalein indicator pink.

to your second question i think that a higher concentration of Copper (ii) ion would increase the rate of ionisation of the nickel electrode which will produce a larger voltage. (pure speculation for this question )

[FreeTheBee]

For the first reaction, look up the possible half reactions. Keep in mind that the solvent can participate as well.

Try writing out the Nernst equation for the second question. It will show how the concentrations affect the potential.

[LHM]

Ohh, thank you very much.

Just wondering, how would the phenolphthalein turn the anode yellow? I thought phenolphthalein only turned pink and clear?

[JadenErius]

it actually isnt the phenolphatelin indicator that turned yellow. Its the iodine that is liberated. it dissolves back into the solution. Iodine + water produces a brown solution (if there is enough dilution, it could produce a yellow solution around the anode)