What is the final mass of lead electorde?

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Answer 1 · 2018-03-21T14:53:00

$sf(9.35color(white)(x)g)$

Lead ions are attracted to the cathode where they are discharged:

$sf(Pb^(2+)+2erarrPb)$

This means that 1 mole of $sf(Pb^(2+))$ requires 2 moles of electrons to form 1 mole of $sf(Pb)$ .

The charge on 1 mole of electrons is $sf(9.649xx10^(4))$ Coulombs. This is termed 1 Faraday ( F ).

We can find the number of Faradays passed hence the moles and mass of lead discharged.

$sf(Q=It)$

$:.$ $sf(Q=1.57xx17xx60=1601.4color(white)(x)C)$

This is equal to $sf(1601.4/(9.649xx10^(4))=0.0166color(white)(x)F)$

$sf(2F)$ will discharge $sf(1)$ mole

$:.$ $sf(1F)$ will discharge $sf(1/2)$ mole

$:.$ $sf(0.0166F)$ will discharge $sf(1/(2)xx0.0166=0.0083)$ mole.

$sf(m=nxxA_r)$

$sf(m=0.0083xx207.2=1.72color(white)(x)g)$

This means the final mass of the elecrode will be:

$sf(7.63+1.72=9.35color(white)(x)g)$