0.02g equivalent of Ag was deposited in an electrolysis experiment. Find the quantity of charge passed. If the same charge is passed through a gold solution, 1.314g of gold is deposited. Find the oxidation state of gold. ( atomic mass of Au =197 amu).
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isha62:
okk..u can find it...bt no need of it...i
the book has asked it and thnx for ur help
okk so multiply 96500 with 0.02
Now I have solved it
u will get the charge
if u know the formula well u should easily solve it
okk gud
I have also asked one more question of same chapter try to answer that one
i have solbed it..bt my pic is not sending
i think some connection problem
Answered by
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0.02 g equivalent of Ag was deposited in electrolysis experiment .
we know,
according to Faraday's second Law ,
1 mole of electrons librates 1 equivalent of matter .
e.g 96500 C charge required to remove 1 equivalent of matter .
so, 0.02g equivalent remove = 0.02 × 96500 = 1930 C
now,
same amount of charge require to remove 1.314 g of gold so,
0.02 g equivalent = given wt/equivalent wt
0.02 = 1.314 /(197/n)
where n is oxidation number of gold .
0.02 = 1.314 × n/197
n = 2.99≈ 3
hence,
oxidation number of gold ( Au) = 3
we know,
according to Faraday's second Law ,
1 mole of electrons librates 1 equivalent of matter .
e.g 96500 C charge required to remove 1 equivalent of matter .
so, 0.02g equivalent remove = 0.02 × 96500 = 1930 C
now,
same amount of charge require to remove 1.314 g of gold so,
0.02 g equivalent = given wt/equivalent wt
0.02 = 1.314 /(197/n)
where n is oxidation number of gold .
0.02 = 1.314 × n/197
n = 2.99≈ 3
hence,
oxidation number of gold ( Au) = 3
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