ElectrochemistryHard

Question

When 12,000 coulombs of electricity is passed through the electrolyte, 3.0 g of a metal of atomic mass 96.5 g/mol is deposited. The electro-valency of the metal cation in the electrolyte is

Options

A.+4
B.+3
C.+2
D.−4

Solution

$n_{eq} = \frac{Q}{F} \Rightarrow \frac{3}{96.5} \times n = \frac{12000}{96500} \Rightarrow n = 4$

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