
When a molten lithium chloride (LiCl) is electrolyzed, lithium metal is formed at the cathode. If current efficiency is 75% then how many grams of lithium are liberated when 1930 C of charge pass through the cell?
(Atomic weight: Li= 7)
A. 0.107
B. 0.120
C. 0.28
D. 0.240
Answer
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Hint: To solve this problem, first find the number of moles in the lithium metal. To find this, use the relation between the efficiency, applied charge and time taken. Substitute the values and get the number of moles in lithium metal. But this amount is in Coulomb so convert it in Faraday. Then multiply this number of moles in Faraday to atomic weight of lithium. This will give the amount of lithium liberated in grams.
Complete step by step answer:
Given: Charge (C)= 1930 C
Current efficiency (E)= 75%= 0.75
Number of moles of the metal which is deposited is given by,
$M= ECt$ …(1)
Where, E is the efficiency
C is the applied charge
t is the time taken
Here, t=1,
So, equation. (1) becomes,
$M=EC$
Substituting values in above equation we get,
$M= 0.75 \times 1930$
$\Rightarrow M=1447.5 C$
1 Faraday is 96500 Coulomb. Thus, the number of moles of the metal deposited is,
$M= \dfrac {1447.5}{96500}$
$\Rightarrow M= 0.015 moles$
Weight of the metal is given by,
$W= M \times Atomic \quad weight$
Substituting values in above equation we get,
$W= 0.015 \times 7$
$\therefore W= 0.105 \quad grams$
$\Rightarrow W \approx 0.105 \quad grams$
Hence, 0.107 grams of lithium are liberated when 1930 C of charge passes through the cell.
So, the correct answer is option A i.e. 0.107.
Note:
When current is passed through molten LiCl i.e. lithium chloride, it decomposes into lithium and chlorine gas. Both lithium and chlorine gas move towards the electrode opposite to their own charge and either loses an electron or gains one. Sodium ${Li}^{+}$ is a positive ion so it moves towards the negative electrode while chlorine ${Cl}^{-}$ is a negative ion so it moves towards the positive electrode. Hence, the lithium metal deposits at the cathode while the chlorine gas is formed at the anode.
Complete step by step answer:
Given: Charge (C)= 1930 C
Current efficiency (E)= 75%= 0.75
Number of moles of the metal which is deposited is given by,
$M= ECt$ …(1)
Where, E is the efficiency
C is the applied charge
t is the time taken
Here, t=1,
So, equation. (1) becomes,
$M=EC$
Substituting values in above equation we get,
$M= 0.75 \times 1930$
$\Rightarrow M=1447.5 C$
1 Faraday is 96500 Coulomb. Thus, the number of moles of the metal deposited is,
$M= \dfrac {1447.5}{96500}$
$\Rightarrow M= 0.015 moles$
Weight of the metal is given by,
$W= M \times Atomic \quad weight$
Substituting values in above equation we get,
$W= 0.015 \times 7$
$\therefore W= 0.105 \quad grams$
$\Rightarrow W \approx 0.105 \quad grams$
Hence, 0.107 grams of lithium are liberated when 1930 C of charge passes through the cell.
So, the correct answer is option A i.e. 0.107.
Note:
When current is passed through molten LiCl i.e. lithium chloride, it decomposes into lithium and chlorine gas. Both lithium and chlorine gas move towards the electrode opposite to their own charge and either loses an electron or gains one. Sodium ${Li}^{+}$ is a positive ion so it moves towards the negative electrode while chlorine ${Cl}^{-}$ is a negative ion so it moves towards the positive electrode. Hence, the lithium metal deposits at the cathode while the chlorine gas is formed at the anode.
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