If the density of copper is 8-94 g-cm 3- the number of Faradays required to plate an area 10 - 10 cm 2 of thickness of 10 - 2 cm using CuSO4 solution as electrolyte is-

The correct option is C 0.28 FVolume of Cu=10× 10× 10^ 2=1 cm^3 Density=MassVolume Mass=8.94 g Cu^2++2e^ ⟶ Cu 63.5 g ...

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Standard XII

Chemistry

Faraday's First Law

If the densit...

Question

If the density of copper is 8.94 g/cm $^{3}$ , the number of Faradays required to plate an area $10 \times 10$ cm $^{2}$ of thickness of $1 0^{- 2}$ cm using $CuS O_{4}$ solution as electrolyte is:

0.1 F

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0.28 F

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0.4 F

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0.5 F

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Solution

The correct option is C 0.28 F
Volume of $C u = 10 \times 10 \times 10^{- 2} = 1$ $c m^{3}$
$D e n s i t y = \frac{M a s s}{V o l u m e}$
$M a s s = 8.94$ $g$
$C u^{2 +} + 2 e^{-} ⟶ C u$
$63.5$ $g ⟶ 2$ $m o l$ $F$
$8.94$ $g ⟶ \frac{2}{63.5} \times 8.94 = 0.28$ $F$

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If the density of copper is 8.94 g/cm3, the number of Faradays required to plate an area 10×10 cm2 of thickness of 10−2 cm using CuSO4 solution as electrolyte is:

The correct option is C 0.28 FVolume of Cu=10×10×10−2=1 cm3Density=MassVolumeMass=8.94 gCu2++2e−⟶Cu63.5 g⟶2 mol F8.94 g⟶263.5×8.94=0.28 F

If the density of copper is 8.94 g/cm $^{3}$ , the number of Faradays required to plate an area $10 \times 10$ cm $^{2}$ of thickness of $1 0^{- 2}$ cm using $CuS O_{4}$ solution as electrolyte is:

The correct option is C 0.28 F
Volume of $C u = 10 \times 10 \times 10^{- 2} = 1$ $c m^{3}$
$D e n s i t y = \frac{M a s s}{V o l u m e}$
$M a s s = 8.94$ $g$
$C u^{2 +} + 2 e^{-} ⟶ C u$
$63.5$ $g ⟶ 2$ $m o l$ $F$
$8.94$ $g ⟶ \frac{2}{63.5} \times 8.94 = 0.28$ $F$