Answer:
1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.
Step-by-step explanation:
According to Faraday's second law of electrolysis, the amount of ions/mass of substance deposited at an electrode depends on its equivalent weight.
For a divalent ion, it will require 2F of electricity per mole.
1 F = 96500 C
Amount of electricity that passes through the electrolyte per second = (magnitude of current) × (time) = It = (1.2 × 1) = 1.2 C
2F (2×96500C) of electricity will deposit 1 mole of Copper ions
That is,
193000 C of electricity will deposit 1 mole of Copper.
1.2 C will deposit (1.2×1/193000); 0.0000062176 mole of Copper.
1 mole of Copper contains (6.022 × 10²³) ions according to the Avogadro's constant.
0.0000062176 mole of Copper will contain (0.0000062176 × 6.022 × 10²³) ions = (3.744 × 10¹⁸) ions.
Therefore, 1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.
Hope this Helps!!!