Question

Calculate the current (in mA) required to deposit 0.195 g of platinum metal in 5 hours from a solution of [PtCl6]₋² : (given at wt of pt =195
1) 310
2) 31
3)21.44
4)5.36

Answer 1

answer : option (3) 21.44

explanation : platinum metal deposits in 5 hours = 0.195 g

atomic mass of platinum = 195 g/mol

so, mole of platinum metal = 0.195/195 = 10^-3 mol

here, you can see oxidation state of platinum is +4 so, To deposit 1 mole of platinum, require 4 mole of electrons.

number of mole of electrons = 4 × 10^-3 mol.

we know, 1 mole of electrons, liberates 1 equivalent of matter.

so, charge flows through plate in 5 hours = 4 × 10^-3 × 96500 = 386 C

now, current = charge/time taken

time taken = 5 hour = 5 × 3600 sec

so, current = 386/(5 × 3600)

= 0.02144 A

= 21.44 × 10^-3 A

= 21.44 mA

hence, answer is option (3)

Answer 2

Answer:

Answer is 21.44

Explanation: