1.

How much Pt is deposited on cathode when 0.80F current is pass through 1.0 M solution of Pt^(4+) ?

Answer»

1.0 MOLE
0.20 mole
0.40 mole
0.80 mole

Solution :`PT^(4+) +4E^(-) to Pt`
`4F=1` mole Pt
`THEREFORE 0.80F=(?)`
`=(1xx0.80)/(4)=0.20` mole.


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