Question

Question #24d99

Answer 1

`390.5"s"` or `"6.5min"`

Explanation:

The first step is to find the quantity of electricity, in Coulombs, to discharge `1.63"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:

`Ni_((aq))^(2+)+2erarrNi_((s))`

`:.` 1 mole `"Ni"^(2+)` needs 2 moles of electrons to produce 1 mole `"Ni"`.

We can find the charge on 1 mole of electrons by multiplying the electronic charge by the Avogadro Constant. This gives `9.64xx10^(4)"C"` and is referred to as the Faraday Constant `"F"`.

`:. 58.7"gNi"^(2+)` require `2"F"` to produce `58.7"gNi"`

`2"F"=2xx9.64xx10^(4)=1.928xx10^(5)"C"`

So:

`58.7"g Ni"` is formed from `1.928xx10^(5)" ""C"`

`:.1"gNi"` is formed from `(1.928xx10^(5))/(58.7)" ""C"`

`:.1.63"gNi"` is formed from `(1.928xx10^(5))/(58.7)xx1.63" ""C"`

`=5.35xx10^(3)" ""C"`

Electric current is the rate of flow of charge:

`I=Q/t`

`:.t=Q/I`

`t=(5.35xx10^(3))/13.7=390.5"s"`

`=390.5/60=6.5"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.