Question

A direct current deposits 54g of silver (atomic mass= 108) during electrolysis. How much aluminium (atomic number = 27) will be deposited by the same amount of electricity?
(A) 4.5g
(B) 5.4g
(C) 54g
(D) 27g

Answer 1

Hint: Find out the oxidation states of silver and aluminium in an electrolyte like silver nitrate and aluminium nitrate. Then, calculate the number of moles of silver metal deposited. From the number of moles, calculate the number of individual silver atoms deposited and then you will obtain the number of electrons present in the whole cell. Now, just do the reverse calculation and obtain a mass of aluminium.

Complete step by step solution:
-Let’s have a look at the given data. A direct current deposits 54g of silver (atomic mass= 108) during electrolysis.
-Number of moles of silver deposited $=\dfrac{Weight}{Molecular\,weight}=\dfrac{54}{108}=0.5\,moles$
-Silver exists in +1 oxidation state in any electrolyte.
-This reaction of metal deposition is a reduction and can be explained as,
\[A{{g}^{+}}+{{e}^{-}}\to Ag\]
-So, for one mole of silver metal to get deposited on the cathode, one mole of electrons is required. One mole of electrons means Avogadro’s number of electrons.
-Here, 0.5 moles of silver are getting deposited so, 0.5moles of electrons are present in the cell. That means, $0.5\times 6.023\times {{10}^{23}}=3.0115\times {{10}^{23}}$ electrons are present.
-Aluminium exists in+3 oxidation state in any electrolyte. That means, 1 mole of aluminium will need 3 moles of electrons to get deposited at the cathode.
\[A{{l}^{3+}}+3{{e}^{-}}\to Al\]
-Now, we know the number of electrons supplied to the cell. Since 1 mole of Al needs 3 moles of electrons, we will have to divide $3.0115\times {{10}^{23}}$electrons by 3 to get the number of molecules of aluminium getting deposited.
-Therefore, $\dfrac{3.0115\times {{10}^{23}}}{3}=1.004\times {{10}^{23}}$molecules of aluminium will get deposited.
-Now, we need to find the number of moles of aluminium. Therefore, $\dfrac{1.004\times {{10}^{23}}}{6.023\times {{10}^{23}}}=0.167\,moles$.
-By multiplying the number of moles with molar mass we get, $0.167\times 27=4.5g$
-Therefore, 4.5g of aluminium will get deposited.

Hence, the correct option is option (A) 4.5g.



Note: Systematically calculate each and every value. Be careful while doing the back-calculation. Remember that one mole is Avogadro’s number of species. Also, to find out mass we need to multiply molar mass and the number of moles of that species.