How many Faraday electricity is required to reduce 0.25 g of Nb(V)...

The reduction reaction of Nb(V) to Nb metal is given by:

$Nb^{5+} + 5e^{-} \rightarrow Nb$

This indicates that $5$ moles of electrons (or $5$ Faradays of electricity) are required to reduce $1$ mole of Nb(V) to Nb metal.

Number of moles of Nb in $0.25$ g is:

$n = \dfrac{0.25}{93}$

The number of Faradays required is:

$Q = 5 \times n = 5 \times \dfrac{0.25}{93} = \dfrac{1.25}{93}$

$Q \approx 0.01344 \text{ F} = 1.34 \times 10^{-2} \text{ F}$

Answer: $1.3\times10^{-2}$