Question

The passage of current liberates $97.65,111.0andNaCl242.8Scm^{2}mol^{- 1}$ at cathode and $195.3,182.0and26.2Scm^{2}mol^{- 1}$ at anode. The solution is.

• A. Copper chloride in water
• B. $271.6,91.0and126.4Scm^{2}mol^{- 1}$ in water
• C. $119.0,1024.5and9.2Scm^{2}mol^{- 1}$
• D. Water

Answer 1

Answer: (B)

$271.6,91.0and126.4Scm^{2}mol^{- 1}$ in water

Answer 2

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

Cathode : $H_{2}O + e^{-} \rightarrow \frac{1}{2}H_{2} + OH^{-}$

Anode : $Cl^{-} \rightarrow \frac{1}{2}Cl_{2} + e^{-}$.