The passage of current liberates H 2 at cathode and Cl 2 at anode. The solution is

Question

The passage of current liberates H 2 at cathode and Cl 2 at anode. The solution is (A) copper chloride in water (B) NaCl in water (C) H 2 SO 4 (D) water.

Tardigrade Admin · Accepted Answer

Since discharge potential of water is greater than that of sodium so, water is reduced at cathode instead of

N a^{+}

.

Cathode

: H_{2} O + e^{-} ⟶ 1/2 H_{2} + O H^{-}

Anode

: C l^{-} ⟶ 1/2 C l_{2} + e^{-}

Q. The passage of current liberates $H_{2}$ at cathode and $C l_{2}$ at anode. The solution is

copper chloride in water

$N a Cl$ in water

$H_{2} S O_{4}$

water.

Since discharge potential of water is greater than that of sodium so, water is reduced at cathode instead of $N a^{+}$ .

Cathode $: H_{2} O + e^{-} ⟶ 1/2 H_{2} + O H^{-}$

Anode $: C l^{-} ⟶ 1/2 C l_{2} + e^{-}$

Q. The passage of current liberates H2​ at cathode and Cl2​ at anode. The solution is

copper chloride in water

NaCl in water

H2​SO4​

water.

Since discharge potential of water is greater than that of sodium so, water is reduced at cathode instead of Na+. Cathode :H2​O+e−⟶1/2H2​+OH− Anode :Cl−⟶1/2Cl2​+e−

Q. The passage of current liberates $H_{2}$ at cathode and $C l_{2}$ at anode. The solution is

$N a Cl$ in water

$H_{2} S O_{4}$

Since discharge potential of water is greater than that of sodium so, water is reduced at cathode instead of $N a^{+}$ .

Cathode $: H_{2} O + e^{-} ⟶ 1/2 H_{2} + O H^{-}$

Anode $: C l^{-} ⟶ 1/2 C l_{2} + e^{-}$