Question

A.How many amperes are required to deposit 0.155 grams of aluminum metal in 507 seconds, from a solution that contains Al³⁺ ions .
_______A.

B.How many amperes are required to deposit 0.283 grams of chromium metal in 303 seconds, from a solution that contains Cr³⁺ ions .
______A.

Answer 1

a)

mol of aluminium = mass/MW = 0.155 / 26.98 = 0.0057449 moles of Al(s) required

so

Al+3 + 3e- --> Al(s)

so we need --> 3 electron per mol

0.0057449 mol of Al = 3*0.0057449 mol of e- = 0.0172347 e-

relate to charge with F constant

F = 96500 C/mol

so

charge = mol of e- * F = 0.0172347*96500 = 1663.148 C

now...

I = C/t =1663.148 /507 = 3.280 amps

B)

similar...

mol of aluminium = mass/MW = 0.283/ 51.9961 = 0.0054427 moles of Al(s) required

so

Cr+3 + 3e- --> Cr(s)

so we need --> 3 electron per mol

0.0054427 mol of Cr = 3*0.0054427mol of e- = 0.016326 e-

relate to charge with F constant

F = 96500 C/mol

so

charge = mol of e- * F = 0.016326*96500 = 1575.459 C

now...

I = C/t =1575.459/303= 5.1995 amps