Two electrolytic cells containing $C u S O_{4}$ and $A g N O_{4}$ respectively are connected in series and a current is passed through them unit 1 mg of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately
[ Atomic weight of copper and silver are respectively 63.57 and 107.88]

1.7 m g

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3.4 m g

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5.1 m g

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6.8 m g

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Solution

The correct option is B $3.4 m g$
$\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}}$ (By faraday law for same current and time)
Where $E_{1}$ and $E_{2}$ are the chemical equivalents and $m_{1}$ and $m_{2}$ are the masses of copper and silver respectively.
$E = \frac{A t o m i c w e i g h t}{V a l e n c y} E_{1} = \frac{63.57}{2} = 31.79$ and
$E_{2} = \frac{107.88}{1} = 107.88$
$∴ \frac{1 m g}{m_{2}} = \frac{31.79}{107.88} \Rightarrow m_{2} = \frac{107.88}{31.79} m g = 3.4 m g$

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The correct option is B 3.4mgm1m2=E1E2 (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=AtomicweightValencyE1=63.572=31.79 and E2=107.881=107.88∴1mgm2=31.79107.88⇒m2=107.8831.79mg=3.4mg