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Standard XII

Chemistry

Charge and Current

The thickness...

Question

The thickness in cm of plating if the cathode consists of a sheet of 4 cm $^{2}$ which is coated on both sides. The density of $N i$ is 8.9 g ml $^{- 1}$ (If the thickness is $X$ cm, then give answer as $100 X$ ).

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Solution

The current efficiency for $N i$ production is 60%.
The number of faradays of electricity passed for $N i$ formation is $\frac{0.6 \times 15 \times 3600}{96500} = 0.3357$ faraday.
The mass of $N i$ deposited is $\frac{0.3357}{2} \times 58.7 = 10 g$
The coated area of the sheet is $2 \times 4 = 8 c m^{2}$ as both sides are coated.
$V o l u m e = \frac{M a s s}{D e n s i t y} = \frac{10}{8.9} = 1.1$
The thickness of the plate is $t h i c k n e s s = \frac{v o l u m e}{a r e a} = \frac{1.1}{8} = 0.14$ cm.
( $100 X = 14$ )

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The thickness in cm of plating if the cathode consists of a sheet of 4 cm2 which is coated on both sides. The density of Ni is 8.9 g ml−1(If the thickness is X cm, then give answer as 100X).

The thickness in cm of plating if the cathode consists of a sheet of 4 cm $^{2}$ which is coated on both sides. The density of $N i$ is 8.9 g ml $^{- 1}$ (If the thickness is $X$ cm, then give answer as $100 X$ ).