Solveeit Logo
Login

Question

Question: In electrolysis of aqueous copper sulphate, the gas at anode and cathode is....

In electrolysis of aqueous copper sulphate, the gas at anode and cathode is.

A

NH4OHNH_{4}OH and NH4OHNH_{4}OH

A

(d) Pb2++2e−→Pb;(Eº=−0.13V)Pb^{2 +} + 2e^{-} \rightarrow Pb;(Eº = - 0.13V) and Ag++e−→Ag;(Eº=+0.80V)Ag^{+} + e^{-} \rightarrow Ag;(Eº = + 0.80V)

B

ΔºNH4OH=ΔºBa(OH)2+ΔºNH4Cl−ΔºBaCl2\Delta º_{NH_{4}OH} = \Delta º_{Ba(OH)_{2}} + \Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}} and ΔºNH4OH=ΔºBaCl2+ΔºNH4Cl−ΔºBa(OH)2\Delta º_{NH_{4}OH} = \Delta º_{BaCl_{2}} + \Delta º_{NH_{4}Cl} - \Delta º_{Ba(OH)_{2}}

C

ΔºNH4OH=ΔºBa(OH)2+2ΔºNH4Cl−ΔºBaCl22\Delta º_{NH_{4}OH} = \frac{\Delta º_{Ba(OH)_{2}} + 2\Delta º_{NH_{4}Cl} - \Delta º_{BaCl_{2}}}{2} and ΔºNH4OH=ΔºNH4Cl+ΔºBa(OH)22\Delta º_{NH_{4}OH} = \frac{\Delta º_{NH_{4}Cl} + \Delta º_{Ba(OH)_{2}}}{2}

Answer

NH4OHNH_{4}OH and NH4OHNH_{4}OH

Explanation

Solution

Cathode : 2H2O+2e−→H2+2OH−2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}

Anode : H2O→2H++12O2+2e−H_{2}O \rightarrow 2H^{+} + \frac{1}{2}O_{2} + 2e^{-}.