Question

To deposit 1 mole of copper from the copper sulphate solution 193000 coulomb charge is required.Find the time to deposit 0.1 mole copper from the copper sulphate solution. If a steady current of 5 amphere is passed through the solution

Answer 1

3860 seconds are required to deposit 0.1 mole copper from the copper sulphate solution.

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

1 mole of copper is deposited by = 193000 C

Thus 0.1 mole of copper is deposited by = $\frac{193000}{1}\times 0.1=19300C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 5 A

q = total charge = 19300 C

t = time required = ?

Putting values in above equation, we get:

$%A=\frac{19300C}{t}\\\\t=\frac{19300}{5A}=3860s$$5A=\frac{19300C}{t}$

$t=\frac{19300C}{5A}=3860s$

Hence, the amount of time needed for 0.1 mole of copper to deposit is 3860 seconds

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