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Kohlrausch Law

In an electro...

Question

In an electroplating experiment , $m$ gram of silver is deposited when $4 A$ current flows for $2 m i n$ . How many gram of silver will be deposited by $6 A$ current flowing for $40 s$ ?

m

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0.5 m

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1.2 m

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2 m

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Solution

The correct option is B $0.5 m$
From Faraday's first law of Electrolysis:-

$\frac{Q}{m} = c o n s t a n t ⟹ \frac{I t}{m} = c o n s t a n t$

Hence, $\frac{I_{2} t_{2}}{m_{2}} = \frac{I_{1} t_{1}}{m_{1}}$

$⟹ \frac{6 \times 40}{m_{2}} = \frac{4 \times 120}{m}$

$⟹ m_{2} = 0.5 m$

Answer-(B)

Suggest Corrections

Similar questions

In an electroplating experiment m g of silver is deposited, when 4 amperes

of current flows for 2 minutes. The amount (in gms) of silver deposited by 6

amperes of current flowing for 40 seconds will be

[MNR 1991]

m g

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25 m i n

1.5 A

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= 3.3 \times 10^{- 4} g C^{- 1},

= 1.1 \times 10^{- 4} g C^{- 1}

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