Question

Determine the mass (g) of aluminum deposited in an electrolytic cell using a current of 0.750 A for 1.2 hour (1 hour = 3,600 s) Al3+ (aq) + 3e - Al(s) 0.302 g 0.000839 g 0.625 g 0.906 g O 2.20 g

Answer 1

Answer : 0.302 g

In electrolysis process,

W = Z * I * t

where, W = mass deposited (g) ; Z = electrochemical equivalent.

Z = Equivalent weight / 96485

where , 96485 C is the amount of electricity passed by one mole of electrons.

and Equivalent weight = molecular weight / valency

Al mol wt = 27 g/mol and valency = 3.

Eq Wt of Al = 27 / 3 = 9

therefore Z = 9 / 96485 = 9.32 * 10^-5

I = amount of current passed (A) and

Time (s) in which current is passed = t = 1.2 hour = 1.2 * 3600 sec = 4320 sec

W = Z*I*t

W = 9.32*10^-5 * 0.750 * 4320

W = 0.30196 gm or

W = 0.302 gm

0.302 gm Aluminium is deposited.