Question

How much silver (in grams) was there in the solution if all the silver was removed as Ag metal by electrolysis for 0.55 hr with a current of 1.10 mA (1 mA = 10-3 A)?

Answer 1

Electrolysis equation is:

Ag1+ + 1e- ------> Ag

1 mol of Ag requires 1 mol of electron

1 mol of electron = 96485 C

So,1 mol of Ag requires 96485 C

let us calculate the charge passed:

t = 0.55 hr = 0.55*3600 s = 1.98*10^3 s

time, t = 1.98*10^3s

Q = I*t

= 1.1*10^-3A * 1.98*10^3s

= 2.178 C

mol of Ag plated = 2.178/96485 = 2.257*10^-5 mol

Molar mass of Ag = 1.079*10^2 g/mol

mass of Ag = number of mol * molar mass

= 2.257*10^-5 * 1.079*10^2

= 2.436*10^-3 g

Answer: 2.44*10^-3 g