How many Faradays are required to reduce 0.25g of Nb (V) to the metal?

Options

A. `2.7xx10^(-3)`
B. `1.3xx10^(-2)`
C. `2.7xx10^(-2)`
D. `7.8xx10^(-3)`

Answer

To determine how many Faradays are required to reduce 0.25 g of Nb (V) to the metal, we can follow these steps:

### Step 1: Determine the atomic mass of niobium (Nb)
The atomic mass of niobium (Nb) is approximately 93 g/mol.

### Step 2: Write the reduction reaction
The reduction of niobium (V) ion (Nb^5+) to niobium metal (Nb) can be represented as:
\[ \text{Nb}^{5+} + 5e^- \rightarrow \text{Nb} \]
This indicates that 5 moles of electrons (or 5 Faradays) are required to reduce 1 mole of Nb^5+ to Nb.

### Step 3: Calculate the number of moles of niobium in 0.25 g
To find the number of moles of niobium in 0.25 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of Nb} = \frac{0.25 \text{ g}}{93 \text{ g/mol}} \approx 0.002688 \text{ mol} \]

### Step 4: Calculate the number of Faradays required
Since it takes 5 Faradays to reduce 1 mole of Nb, we can calculate the Faradays needed for 0.002688 moles:
\[ \text{Faradays required} = \text{Number of moles} \times 5 \]
Substituting the number of moles:
\[ \text{Faradays required} = 0.002688 \text{ mol} \times 5 \approx 0.01344 \text{ Faradays} \]

### Step 5: Final result
Thus, the number of Faradays required to reduce 0.25 g of Nb (V) to the metal is approximately 0.01344 Faradays.

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