An aluminium foil with a surface area of 500cm2 was anodized. Gow many coulombs of electricity were required to increase the thickness of the layer by 1*10^-3 cm (density of aluminium oxide=4.0g cm-3)?

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Answer 1 · 2018-01-28T16:40:52

11,000 C

Anodising is a process by which the protective oxide layer on the aluminium is thickened by an electrolytic process.

Sodium hydroxide solution is used as the electrolyte. At the anode $sf(OH^-)$ are oxidised:

$sf(4OH^(-)rarrH_2O+O_2+4e)$

The oxygen reacts with the aluminium:

$sf(2Al+3/2O_2rarrAl_2O_3)$

The 1/2 equations are:

$sf(2Alrarr2Al^(3+)+6e)$

$sf(3/2O_2+6erarr3O^(2-))$

In reality you can use a.c current and coat 2 pieces of aluminium.

You can see that to form 1 mole of $sf(Al_2O_3)$ you need 6 moles of electrons.

The charge carried by 1 mole of electrons is referred to as The Faraday Constant F which is $sf(9.65xx10^(4)"C/mol")$

We can use the date given to find the actual no. of moles deposited.

$sf(m=rhoxxv)$

$sf(v=500xx10^(-3)=0.5color(white)(x)cm^3)$

$:.$ $sf(m=4.0xx0.5=2.0color(white)(x)g)$

$sf(M_r[Al_2O_3]=102)$

$:.$ $sf(n_(Al_2O_3)=m/M_r=2.0/102=0.0196)$

We know that:

1 mole requires 6F Coulomb.

$:.$ 0.0196 mole requires 0.0196 x 6F = 0.1176 F Coulomb

$sf(=0.1176xx9.65xx10^(4)=11348color(white)(x)C)$

$sf(~~11,000color(white)(x)C)$