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Question: One mole of electron passes through each of the solution of AgNO$_3$, CuSO$_4$, and AlCl$_3$ when Ag...

One mole of electron passes through each of the solution of AgNO3_3, CuSO4_4, and AlCl3_3 when Ag, Cu and Al are deposited at cathode. The molar ratio of Ag, Cu and Al deposited are

A

1:1:1

B

6:3:2

C

6:3:1

D

1:3:6

Answer

The molar ratio of Ag, Cu and Al deposited are 6:3:2.

Explanation

Solution

The deposition of metals at the cathode occurs via reduction reactions, where metal ions gain electrons. The number of electrons required to deposit one mole of a metal depends on the charge of its ion.

  1. For Silver (Ag): The ion is Ag+^+.

    The reduction reaction is: Ag++e−→Ag(s)\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag(s)}

    This indicates that 1 mole of electrons deposits 1 mole of Ag.

    Since 1 mole of electrons passes through the solution, moles of Ag deposited = 1 mole.

  2. For Copper (Cu): The ion is Cu2+^{2+}.

    The reduction reaction is: Cu2++2e−→Cu(s)\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu(s)}

    This indicates that 2 moles of electrons deposit 1 mole of Cu.

    Since 1 mole of electrons passes through the solution, moles of Cu deposited = 12\frac{1}{2} mole = 0.5 mole.

  3. For Aluminum (Al): The ion is Al3+^{3+}.

    The reduction reaction is: Al3++3e−→Al(s)\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al(s)}

    This indicates that 3 moles of electrons deposit 1 mole of Al.

    Since 1 mole of electrons passes through the solution, moles of Al deposited = 13\frac{1}{3} mole.

The molar ratio of Ag : Cu : Al deposited is therefore: 1:0.5:131 : 0.5 : \frac{1}{3}

To express this ratio in whole numbers, we multiply each term by the least common multiple (LCM) of the denominators (1, 2, 3), which is 6. (1×6):(0.5×6):(13×6)(1 \times 6) : (0.5 \times 6) : \left(\frac{1}{3} \times 6\right) 6:3:26 : 3 : 2

Thus, the molar ratio of Ag, Cu, and Al deposited is 6:3:2.