ElectrochemistryHard
Question
From an electrolyte, one mole of electron will deposit at cathode
Options
A.63.5 g of Cu
B.24 g of Mg
C.11.5 g of Na
D.9.0 g of Al
Solution
$n_{eq}Cu = \frac{63.5}{63.5} \times 2 = 2;n_{eq}Mg = \frac{24}{24} \times 2 = 2$
$n_{eq}Na = \frac{11.5}{23} \times 1 = 0.5;n_{eq}Al = \frac{9}{27} \times 3 = 1$
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