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Question

The number of electrons participating in the electrode reaction when one atomic weight of a bivalent metal was deposited at the cathode:


  1. 9.65 x 1023
  2. 12.04 x 1023
  3. 0.6 x 1023
  4. 3.01 x 1023

A
12.04 x 1023
B
9.65 x 1023
C
0.6 x 1023
D
3.01 x 1023
Solution
Verified by Toppr

Let, Mg2+ be the bivalent metal ion.

Now,
At cathode: Mg2++2eMg(s)

So, 2 moles of electrons is required to obtain 1 mol of metal atoms.

So, number of electrons required =2×6.02×1023=12.04×1023

Option C is correct.

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