Calculating the amount of charge from the amount of product in an electrolysis: How many amperes must be passed through a Downs cell to produce sodium metal a rate of `30.0 Kg//h.`
Strategy : Producted through a sequence of conversion similar to that in worked Example 3.26 but in reverse order
`Na^(+1) + e^(-) rarr Na`
The electrode equation for sodium says that `1` mol `Na` is equiva-lent to `1` mole `e^(-)`. We can use this in the conversion of grams of `Na`. The Faraday constant (which says that one mole of electrons is equivalent to `9.65 xx 10^(4)C`) converts mole of electrons to cou-lombs. The conversions are
`g Na rarr mol Na rarr mol e^(-) rarr "coulombs" (C)`
The current in ampres `(A)` equals the charge in coulombs divided by the time in seconds.

Explanation

Because the moalr mass of sodium is `23.0 g//mol`, the number of moles of sodium produced per hour is
Moles of `Na = (30.0 Kg Na) ((1000 g)/(1 Kg)) (1 mol Na)/(23.0 gNa))`
`= 1304.3 mol Na`
` 1.30 xx 10^(3) mol Na`
To produce each mole of sodium, `1` mol of electrons must be passed through the cell :
`Na^(+)(l) + e^(-) rarr Na(l)`
Therefore, the charge passed per hour is
Charge `=(1.30xx10^(3) "mol Na")((1mol e^(-))/(1 mol Na))((96,500)/(1 mol e^(-)))`
`= 125450 xx 10^(3)C`
`= 1.25 xx 10^(8)C`
Since there are `3600` s in `1` h, the current required is
Current `= (1.25 xx 10^(8)C)/(3600S)`
`= 0.000347 xx 10^(8)C/s`
`= 3.47 xx 10^(4)C/s`
`= 34, 700 A`
As a shortcut, the entire conversion of grams of `NA` to coulombs required to deposit this amount to sodium is `(30.0 Kg)((1000 g)/(1 Kg))((1 mol Na)/(23.0 g Na))((1 mol e^(-))/(23.0 g Na)) ((96,500 C)/(1 mol e^(-)))`
`= 1.25 xx 10^(8)C`