Give the products of electrolysis in each case.
(Also, give ionic equation)
Concentrated $N a C l$ solution using Platinum electrodes.

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Solution

Electrolysis is the process of breaking the substance into its constituent parts by passing electricity through it when dissolved in water.
Step 1: Reaction involved
The reaction involved is written below
$N a C l (a q) \to N a^{+} (a q) + C l^{-} (a q) H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q)$
Step 2: Reaction at the Cathode
The concentration of $N a^{+}$ ions is more than that of $H^{+}$ ions. So, the reaction at the Cathode will be
$N a^{+} (a q) + e^{-} \to N a (s)$
Step 3: Reaction at the Anode
The concentration of $C l^{-}$ ions is more than that of $O H^{-}$ ions. So, the reaction at the anode will be
$2 C l^{-} (a q) - 2 e^{-} \to 2 C l (g) C l (g) + C l (g) \to C l_{2} (g)$
So, Sodium $(N a)$ is produced at the cathode and Chlorine $(C l_{2})$ will be produced at the anode.

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