Question

A solution of Cu(NO₃)₂ is electrolyzed between platinum electrodes using 0.1 Faraday electricity. How many moles of Cu will be deposited at the cathode?

पर्याय

• 0.20

• 0.15

• 0.10

• 0.05

Answer 1

0.05

Explanation:

At cathode:

\[\ce{Cu^{2+} + 2e^- -> Cu}\]

∴ 2F will deposit 1 mole of Cu.

∴ 0.1 F will deposit = `(0.1 xx 1)/2` = 0.05 mol Cu