The same quantity of electricity is passed through `Al_(2)(SO_(4))_(3)` and `AgNO_(3)` solution with platinum electrodes. If `n` number of `Al` atoms are deposited on the cathode, `3n` number of `Ag` atoms will be deposited on the cathode.

Answer

T

Explanation

`1F=96500C=1 Eq of Al^(3+)=1Eq of Ag^(o+)`
`=(1)/(3) mol of Al^(3+)-=(1)/(1) mol of Al^(o+)`
`=1 mol of Al^(3+)=3 mol of Ag^(o+)`
`=n ` atoms of `Al^(3+) = 3n` atoms of `Ag^(o+)`