The number of electrons participating in the electrode reaction when one atomic weight of a bivalent metal was deposited at the cathode:
9.65 x 1023
12.04 x 1023
0.6 x 1023
3.01 x 1023
A
12.04 x 1023
B
9.65 x 1023
C
0.6 x 1023
D
3.01 x 1023
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Solution
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Let, Mg2+ be the bivalent metal ion.
Now,
At cathode: Mg2++2e−→Mg(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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