Consider the reaction, M^{3+}+3e^-\rightarrow M; If 0.415 g of metal M is deposited at the cathode due to passage of 2A of current for 965 sec.
Consider the reaction \(M^{3+}+3e^-\rightarrow M;\) If 0.415 g of metal M is deposited at the cathode due to passage of 2A of current for 965 sec. then the atomic weight of the metal M is(a) 124.5 g(b) 62.25 g(c) 18.75 g(d) 27.5 g
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The answer is (b) 62.25 g. This can be calculated using the formula, \(mass \ of \ metal \ deposited \ = \ current \ x \ time \ x \ Faraday's \ constant/valency \ of \ metal \\ mass \ deposited \ = \ 2A \times 965\ seconds \times 96485C/mol \times 3 \ = \ 0.415 \ g \ or \ 415 \ mg\). Since 1 mole of M has a mass of 62.25 g, 415 mg of M has a mass of 62.25 g.
