One Faraday of electricity is passed through molten Al2O3 aqueous solution of CuSO4 and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al,Cu and Na deposited at the respective cathode is

Question

One Faraday of electricity is passed through molten Al2O3 aqueous solution of CuSO4 and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al,Cu and Na deposited at the respective cathode is (A) 2: 3: 6 (B) 6: 2: 3 (C) 6: 3: 2 (D) 1: 2: 3

Tardigrade Admin · Accepted Answer

Al3+ + 3e- arrow Al Cu2+ + 2e- arrow Cu Na+ + e- arrow Na Hence, 1 F will deposit 1/3 mole of Al, 1/2 mol of Cu and 1 mole of Na ∴ molar ratio = (1/3): (1/2): 1 = 2: 3: 6

Q. One Faraday of electricity is passed through molten $A l_{2} O_{3}$ aqueous solution of $C u S O_{4}$ and molten $N a Cl$ taken in three different electrolytic cells connected in series. The mole ratio of $A l, C u$ and $N a$ deposited at the respective cathode is

$2 : 3 : 6$

$6 : 2 : 3$

$6 : 3 : 2$

$1 : 2 : 3$

$3 : 6 : 2$

$A l^{3 +} + 3 e^{-} \to A l$
$C u^{2 +} + 2 e^{-} \to C u$
$N a^{+} + e^{-} \to N a$
Hence, $1 F$ will deposit $1/3$ mole of $A l$ , $1/2$ mol
of $C u$ and $1$ mole of $N a$
$∴$ molar ratio $= \frac{1}{3} : \frac{1}{2} : 1$
$= 2 : 3 : 6$

Q. One Faraday of electricity is passed through molten Al2​O3​ aqueous solution of CuSO4​ and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al,Cu and Na deposited at the respective cathode is

2:3:6

6:2:3

6:3:2

1:2:3

3:6:2

Al3++3e−→Al Cu2++2e−→Cu Na++e−→Na Hence, 1F will deposit 1/3 mole of Al, 1/2 mol of Cu and 1 mole of Na ∴ molar ratio =31​:21​:1 =2:3:6