Single Choice

If three faradays of electricity is passed through the solution of $$AgNO_3, CuSO_4$$ and $$AuCl_3$$, the molar ratio of the cations deposited at the cathodes will be:

A$$1 :1 : 1$$
B$$1 : 2 : 3$$
C$$3 : 2 :1$$
D$$6 : 3 : 2$$
Correct Answer

Solution

$$\displaystyle AgNO_3 + e^- \rightarrow Ag + NO_3^-$$
$$\displaystyle CuSO_4 + 2e^- \rightarrow Cu + SO_4^{2-}$$
$$\displaystyle AuCl_3 + 3e^- \rightarrow Au + 3Cl^-$$
If three faradays of electricity is passed through the solution of
$$\displaystyle AgNO_3, CuSO_4 $$ and $$\displaystyle AuCl_3$$
$$\displaystyle 3 $$ moles, $$\displaystyle \dfrac {3}{2}$$ moles and $$\displaystyle 1 $$ mole of the cations will be deposited at the cathodes .
The molar ratio of the cations deposited at the cathodes will be $$\displaystyle 3 : \dfrac {3}{2} : 1 $$ or
$$\displaystyle 6 : 3 : 2 $$

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