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Q.

How many Faradays are required to reduce 0.25 g of Nb (v) to the metal?(Atomic weight of Nb= 93)

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a

$2.7 \times 10^{- 2}$

b

$7.8 \times 10^{- 3}$

c

$2.7 \times 10^{- 3}$

d

$1.3 \times 10^{- 2}$

answer is B.

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Detailed Solution

$N b^{+ 5} \to N b$ No. of $e^{-}$ involved $= 5 n = \frac{0.25}{93} = 0.0026$ no of faradays required $= 0.0026 \times 5 = 0.013$

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