What mass of I2 from aqueous KI can be produced in 1.0 hour with a current of 15A?

Question

The part that I am having a problem with is the reaction between I2 and KI. I know KI is a reactant and I2 is a product, and that there must be electrons somewhere in the reaction, but I can't figure out how all that fits in.

Answer 1

I- is being oxidised at the electrode:

2 I- = I2 + 2e-

Something is being reduced at the other electrode, but it doesn't matter what for this problem.

To finish the problem,

time (s) x current (A) = amount of electricity (C)

Moles electricity = number of Faradays = Coulombs/96,500

A Faraday is just a mole of electrons.

Knowing how many Faradays you have, divide by two (2 electrons in the reaction) and that will tell you how many moles I2 you have made.

Principles: this is just an amount balancing problem like any other. So you need to go through moles. The moles and electricity are the Faradays. You get the amount of electricity in C by multiplying time (s) by rate (current) (A).

Hope this helps.