Question

How many faradays are required to reduce 0.25 gram of nb to the metal?

Answer 1

Answer: The number of faradays required is $1.3\times 10^{-2}F$

Explanation:

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

The chemical equation for the reduction of niobium (V) ion to a neutral metal follows:

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

We know that:

Molar mass of niobium = 93 g/mol

We are given:

Mass of niobium (V) ion = 0.25 g

To calculate the amount of faradays required, we use unitary method:

For 93 g of niobium (V) ion, the amount of faradays required are 5 F

So, for 0.25 g of niobium (V) ion, the amount of faradays required will be = $\frac{5F}{93g}\times 0.25g=1.3\times 10^{-2}F$

Hence, the number of faradays required is $1.3\times 10^{-2}F$