# Question #24d99

Apr 21, 2016

$390.5 \text{s}$ or $\text{6.5min}$

#### Explanation:

The first step is to find the quantity of electricity, in Coulombs, to discharge $1.63 \text{g}$.

Once we know this we can find the time this will take since we are told the current, which is the rate of flow of charge.

Nickel(II) ions are discharged:

$N {i}_{\left(a q\right)}^{2 +} + 2 e \rightarrow N {i}_{\left(s\right)}$

$\therefore$ 1 mole ${\text{Ni}}^{2 +}$ needs 2 moles of electrons to produce 1 mole $\text{Ni}$.

We can find the charge on 1 mole of electrons by multiplying the electronic charge by the Avogadro Constant. This gives $9.64 \times {10}^{4} \text{C}$ and is referred to as the Faraday Constant $\text{F}$.

$\therefore 58.7 {\text{gNi}}^{2 +}$ require $2 \text{F}$ to produce $58.7 \text{gNi}$

$2 \text{F"=2xx9.64xx10^(4)=1.928xx10^(5)"C}$

So:

$58.7 \text{g Ni}$ is formed from $1.928 \times {10}^{5} \text{ ""C}$

$\therefore 1 \text{gNi}$ is formed from $\frac{1.928 \times {10}^{5}}{58.7} \text{ ""C}$

$\therefore 1.63 \text{gNi}$ is formed from $\frac{1.928 \times {10}^{5}}{58.7} \times 1.63 \text{ ""C}$

$= 5.35 \times {10}^{3} \text{ ""C}$

Electric current is the rate of flow of charge:

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

$\therefore t = \frac{Q}{I}$

$t = \frac{5.35 \times {10}^{3}}{13.7} = 390.5 \text{s}$

$= \frac{390.5}{60} = 6.5 \text{min}$

In reality a current of this size, as well as being very dangerous, would not give a very good result.

The quality of the deposit at the cathode is favoured by a low current density which is current/area. So a low current over a high area gives a deposit that binds well to the surface.

A high current density like this would give a very spongy deposit which would fall off the cathode very easily.