The same amount of electricity was passed through two cells containing molten Al2O3 and molten NaCl . If 1.8 g of Al were liberated in one cell, the amount of Na liberated in other cell is

Question

The same amount of electricity was passed through two cells containing molten Al2O3 and molten NaCl . If 1.8 g of Al were liberated in one cell, the amount of Na liberated in other cell is (A) 2.8 g (B) 3.2 g (C) 4.6 g (D) 6.8 g

Tardigrade Admin · Accepted Answer

Key Idea According to Faraday second law of electrolysis

W \propto E

\frac{W _{1}}{W _{2}} = \frac{E _{1}}{E _{2}}

Equivalent weight

= \frac{atomic weight}{valency}

W_{A l} = 1.8 g

E_{N a} = 23

E_{N} = \frac{27}{3} = 9

∵ \frac{W _{N a}}{W _{N}} = \frac{E _{N a}}{E _{A l}}

∴ \frac{W _{N a}}{1.8} = \frac{23}{9}

W_{N a} = 4.6 g

Q. The same amount of electricity was passed through two cells containing molten Al2​O3​ and molten NaCl . If 1.8g of Al were liberated in one cell, the amount of Na liberated in other cell is

2.8 g

3.2 g

4.6 g

6.8 g

Key Idea According to Faraday second law of electrolysis W∝E W2​W1​​=E2​E1​​ Equivalent weight = valency atomic weight ​ WAl​=1.8g ENa​=23 EN​=327​=9 ∵WN​WNa​​=EAl​ENa​​ ∴1.8WNa​​=923​ WNa​=4.6g

Q. The same amount of electricity was passed through two cells containing molten $A l_{2} O_{3}$ and molten $N a Cl$ . If $1.8 g$ of $A l$ were liberated in one cell, the amount of $N a$ liberated in other cell is