Calculate the mass of Magnesium that will be liberated from its salt by the same…

Explanation

Firstly,    Ag\(^+\)  +   e\(^-\)  →  Ag

              1 mole = 1 F = 96,500C x 1 e\(^-\)

⇒   If  96,500C liberates  108g of Ag

            x       will  liberate  16g of Ag

           x = \(\frac{96,500 x 16}{108}\)  =  14,296.30 C

i.e The quantity of electricity that would liberate 16g of Ag which is also the same quantity that will liberate X of Mg = 14,296.30 C

Thus,     Mg\(^{2+}\)  +   2e\(^-\)  →  Mg

     If     2e\(^-\)  X  96,500C  liberates 24g of Mg

            14,296.30C  will  liberate    Xg of Mg

By simple proportion, X = \(\frac{14,296.30 x 24}{193,000}\) 

                                      = 1.78g

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