Question

guys:-

solve this question please

----> detailed sol required:-

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Answer 1

ii think opt is a
hey mate cucl2 only can liberate h2 at cathode and cl2 at anode
according to eliminating opts that opt c suits mate... delete this ans if it's surely wrong broo

Answer 2

$\huge{NaCl in water/:}$

$<b>Nacl + H2 gives..Na + OH +1/2 H2 + 1/2 Cl2..$

Sodium chloride and water ionize as follows:

At cathode: Both Na+ and H+ ions 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.

Thus H2 gas is liberated at the cathode and Na+ ions remain in the solution.
At the anode: Both Cl– and OH– ions are present near the anode. As the discharge potential of Cl– ions is lower than that of OH- ions, so Cl– ions are discharged in presence to OH– ions.



Thus Cl2 is liberated at anode and OH– ions remain in the solution.
The overall reaction is:
NaCl(aq) + H2 O(l) → Na+ (aq) + OH– (aq)


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