When dilute sodium chloride is electrolysed using graphite electrodes, the cation which is discharged at the cathode most readily. $[N a^{1 +}, O H^{1 -}, H^{1 +}, C l^{1 -}]$

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Solution

Correct answer : $H^{1 -}$

At cathode:
Both $N a^{+} a n d H^{+}$ are present near the cathode. but the discharge potential of $H^{+}$ is lower than that of $N a^{+}$ ion.so, $H^{+}$ ions are discharged in preference to $N a^{+}$ ions.
$H^{+} + e^{-} \to H$
$H + H \to H_{2}$

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Strength of Acids and Bases

Standard X Chemistry

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When dilute sodium chloride is electrolysed using graphite electrodes, the cation which is discharged at the cathode most readily. [Na1+,OH1−,H1+,Cl1−]

Correct answer :H1− At cathode: Both Na+andH+are present near the cathode. but the discharge potential of H+is lower than that of Na+ion.so,H+ ions are discharged in preference to Na+ions. H++e−→H H+H→H2

When dilute sodium chloride is electrolysed using graphite electrodes, the cation which is discharged at the cathode most readily. $[N a^{1 +}, O H^{1 -}, H^{1 +}, C l^{1 -}]$

Correct answer : $H^{1 -}$

At cathode:
Both $N a^{+} a n d H^{+}$ are present near the cathode. but the discharge potential of $H^{+}$ is lower than that of $N a^{+}$ ion.so, $H^{+}$ ions are discharged in preference to $N a^{+}$ ions.
$H^{+} + e^{-} \to H$
$H + H \to H_{2}$