a current is passed through 500 mL of a solution of CaI2. the following electrode reactions occur:
anode: 2I- ---> I2 + 2e-
cathode: 2H2O + 2e- ---> H2 + 2OH-
after some time, analysis of the solution shows that 35.7 mmol of I2 has been formed.
a) how many faradays of charge have passed through the solution?
b) how many coulombs?
c) what volume of dry H2 at STP has been formed?
d) what is the pH of the solution?
chemistry - DrBob222, Monday, June 18, 2012 at 4:34pm
1F will deposit 1 eq I2 or 253.8/2 = 126.9 g I2.
35.7 mmol = 0.0357 mol or 0.0357 x 2 = 0.0714 equivalents.
1F x 0.0714 eq/F = 0.0714 F
There are 96,485 C in 1 F; therefore,
0.0714 x 96,485 = ?C
For 35.7 mmol I2 there must be 35.7 mmol H2. You don't have a T listed so I don't know how you correct for the vapor pressure of H2O.