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Dear Student,

The cathode reaction in the cell are:

1) Ag⁺ (aq) + 1e^- ---> Ag(s)

2) Cu^{2 +} (aq)  + 2e^-   -----> Cu(s)

3) Al³⁺ (aq) + 3e^- ----> Al (s)

Hence, moles of Ag deposited = 1 x 6 = 6

moles of Cu deposited = (1/2) x 6 = 3

moles of Al deposited = (1/3) x 6 = 2

Molar ratio of Ag : Cu : Al is  6 : 3 : 2


So, the correct option is C) 6 : 3 : 2