Question

How many Faradays are required to reduce \(0.25 \mathrm{~g}\) of \(\mathrm{Nb}(\mathrm{V})\) to the metal? (Atomic weight : \(\mathrm{Nb}=93 \mathrm{~g}\) ) (a) \(2.7 \times 10^{-3}\) (b) \(1.3 \times 10^{-2}\) (c) \(2.7 \times 10^{-2}\) (d) \(7.8 \times 10^{-3}\)

Short answer

The number of Faradays required is approximately 1.3 x 10^-2.

Step-by-step solution

Calculate moles of Niobium (Nb)

To find the moles of niobium, use the formula: moles = mass (g) / atomic weight (g/mol). Given that the mass of Nb is 0.25 g and its atomic weight is 93 g/mol, the moles of Nb is calculated as: moles = 0.25 g / 93 g/mol.

Calculate the amount of Faradays needed

Since Nb is in the +5 oxidation state (V), it requires 5 moles of electrons to reduce to Nb metal. A single Faraday is the charge of one mole of electrons, approximately 96,485 coulombs. Multiply the moles of Nb from Step 1 by 5 to find the number of moles of electrons needed, and then divide by Avogadro's number (6.022 x 10^23 mol^-1) to find the number of Faradays.

Compute the final result

Using the values from the previous steps, calculate the number of Faradays required to reduce 0.25 g of Nb(V) to the metal. The calculation is as follows: number of Faradays = (moles of Nb) * (5 moles of electrons per mole of Nb) / (1 mole of electrons per Faraday).

Key concepts

Moles to Faradays Conversion

When working with electrochemistry and redox reactions, it's often necessary to convert between moles of a substance and the Faradays required to react that substance. Faraday's constant, approximately 96,485 coulombs, represents the charge of one mole of electrons.

To perform this conversion, we simply recognize that each mole of electrons is one Faraday. For example, if we determine that a reaction requires 3 moles of electrons, we can directly say that it requires 3 Faradays. In the case of reducing niobium (Nb) from its +5 oxidation state to the metal, the Nb atom accepts five electrons. Therefore, we need to convert the moles of Nb to moles of electrons and then to Faradays. If we have 0.001 moles of Nb, then we would need 0.005 moles of electrons, which translates to 0.005 Faradays of charge.

Electrochemical Reduction

Electrochemical reduction involves the gain of electrons by a substance during a chemical reaction, which commonly takes place in an electrochemical cell. It's a key element in the conversion of various oxidation states of elements to their metallic form. In our example with niobium (Nb), this process occurs when Nb in its +5 oxidation state gains electrons to form Nb metal.

This type of reaction is used in various applications, including refining metals and in batteries. The number of Faradays corresponds to the amount of electric charge needed to drive this electron transfer. Understanding the relationship between electron flow and chemical change is crucial, thus connecting the stoichiometry of the reaction with the electric current passed through the system.

Stoichiometry in Redox Reactions

Stoichiometry in redox reactions refers to the quantitative relationship between the amount of reactants and products in a chemical reaction involving the transfer of electrons. The term 'redox' stands for reduction-oxidation reactions.

Key to mastering stoichiometry is balance: every electron lost by one species in the reaction must be gained by another. This principle ensures that the reaction adheres to the law of conservation of mass and charge. In our niobium example, the stoichiometry dictates that five moles of electrons are required for each mole of niobium to be reduced from a +5 oxidation state to its metallic form. Understanding these ratios allows us to calculate the required Faradays for the electrochemical reaction.

Niobium Oxidation States

Niobium (Nb) is a transition metal capable of existing in multiple oxidation states. The most common oxidation states for niobium range from +1 to +5, with the +5 oxidation state being the most stable and prevalent, especially in aqueous solutions.

The oxidation state indicates the degree of oxidation of an atom in a compound and is represented by a positive or negative number. In the context of our problem, niobium is initially in the +5 oxidation state, meaning that it will require five electrons to reduce it to its metallic state. This information is the foundation for calculating the moles of electrons and therefore the Faradays required for the full reduction of niobium in an electrochemical process.