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.

The time required for a constant current of 0.868 A to deposit 0.387 g of Tl(III) as Tl(s) is 2689 s

We'll begin by calculating the quantity of electricity needed to deposit 0.387 g of Ti

Ti³⁺ + 3e —> Ti

Recall:

1 mole of Ti = 48 g

1 electron (e) = 96500 C

Thus,

3 electrons = 3 × 96500 = 289500 C

From the balanced equation above,

48 g of Ti was deposited by 289500 C of electricity.

Therefore,

0.387 g of Ti will be deposited by = (0.387 × 289500) / 48 = 2334.09375 C of electricity

Quantity of electricity (Q) = 2334.09375 C

Current (I) = 0.868 A

t = Q / I

t = 2334.09375 / 0.868

Therefore, the time required for the reaction is 2689 s

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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

h let me think

it is a solid

explanation: