Question

Be sure to answer all parts. When an aqueous solution containing gold (III) salt is electrolyzed, metallic gold is deposited at the cathode, and oxygen gas is generated at the anode. (a) If 6.05 g of Au is deposited at the cathode, calculate the volume of 02 generated at 23°C and 752 mmHg. (b) What is the current used if the electrolytic process took 4.30 h?

Answer 1

a) no of mol of Au deposited = 6.05/197 = 0.0307 mol

to deposit 0.0307 mol of Au , 0.0307*3 mol of Electrons are required.

we know that, 1 mol O2 = 4 mol electrons

so that, no of mol of O2 deposited = 0.0307*3/4 = 0.023 mol

volume of O2 deposited (V) = nRT/P

                            = 0.023*0.0821*(23+273.15)/(752/760)

                            = 0.565 L

b) faradays first law

equivalent weight of Au (E) = atwt / charge = 197/3 = 65.67 g/equiv

W = Zit

Z = E/F

w = E/Fit

F = faraday = 96500 C

i = current = ? A

t = time = 4.3*60*60 sec

w = weight of Au deposited = 6.05 g

6.05 = (65.67/96500)*x*4.3*60*60

i = 0.574 amp