Question

Calculate the time required for a constant current of 0.868 A to deposit 0.387 g of Tl(III) as Tl(s) on a cathode.

Answer 1

Answer:

632.7 secs

Explanation:

Equation of the reaction:

Tl^3+ + 3e ------> Tl

If 1F = 96500 C

Q= It

Where;

I= current

t= time

3× 96500 C deposited 204 g of Tl

0.868 × t deposited 0.387 g of Tl

3 × 96500 × 0.387 = 204 × 0.868 × t

t= 3 × 96500 × 0.387/ 204 × 0.868

t= 632.7 secs