A) 1.7 mg
B) 3.4 mg
C) 5.1 mg
D) 6.8 mg
Correct Answer: B
Solution :
\[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\] (By faraday law for same current and time) Where E1 and E2 are the chemical equivalents and m1 and m2 are the masses of copper and silver respectively. \[E=\frac{\text{Atomic weight}}{\text{Valency}}\]. \[{{E}_{1}}=\frac{63.57}{2}=31.79\] and \[{{E}_{2}}=\frac{107.88}{1}=107.88\] \ \[\frac{1\,mg}{{{m}_{2}}}=\frac{31.79}{107.88}\] Þ \[{{m}_{2}}=\frac{107.88}{31.79}mg=3.4\,mg\]
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