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A current of 40 microampere is passed th...

A current of 40 microampere is passed through silver nitrate solution for 16 minutes using platinum electrodes.
50% of the cathode is occupied by a single atom thick silver layer. Calculate the total surface area of the cathode if one silver atom occupies `5.5 xx 10^(-16) cm^(2)` surface area.

Text Solution

Verified by Experts

Mass of silver deposite may be calculated according to Faraday's first law of electrolysis.
`W = (I tE)/(96500)`
`= (40 xx 10^(-6) xx 60 xx 16 xx 108)/(96500)`
`= 42.976 xx 10^(-6) g`
Total number of deposited 'Ag' atoms
`= (42.976 xx 10^(-6))/(108) xx 6.023 xx 10^(23)`
`= 2.3967 xx 10^(17)` atoms
Surface occupied by deposited silver
= Number of silver atoms `xx` Area occupied by a single atom
`= 2.3967 xx 10^(17) xx 5.5 xx 10^(-16)`
`= 131.818 cm^(2)`
Since, deposited silver occupies 50% of total area of cathode, hence, total surface area of cathode `= 2 xx 131.818`
`= 263.636 cm^(2)`
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