The electrolysis of \(\mathrm{BiO}^{+}\) produces pure bismuth. How long would it take to produce \(10.0 \mathrm{g}\) Bi by the electrolysis of a \(\mathrm{BiO}^{+}\) solution using a current of \(25.0 \mathrm{A} ?\)

First, we need to find the moles of bismuth (\(n_{Bi}\)) to be produced. To do this, divide the mass of bismuth (\(m_{Bi} = 10 g\)) by its molar mass (\(M_{Bi} = 208.98 g/mol\)): \[n_{Bi} = \frac{m_{Bi}}{M_{Bi}}\]

In the electrolysis process, BiO⁺ ion loses one electron to form pure bismuth (Bi). Therefore, one mole of bismuth requires one mole of electrons. So, the moles of electrons needed (\(n_{e}\)) are equal to the moles of bismuth: \[n_{e} = n_{Bi}\]

Now, we need to find the total charge (in coulombs) required for the electrolysis process (\(Q\)). Faraday's constant (\(F\)) states that one mole of electrons carries a charge of 96,485 C/mol. Therefore, to find the charge needed, multiply the moles of electrons by Faraday's constant: \[Q = n_{e} \times F\]

Finally, to find the time required for electrolysis (\(t\)), we will use the formula for electric current (\(I\)), which is the charge per unit time: \[I = \frac{Q}{t}\] Rearranging the equation to find the time, we get: \[t = \frac{Q}{I}\] Now, let's find the time by plugging in the given values: 1. Calculate the moles of bismuth to be produced: \[n_{Bi} = \frac{10.0 \mathrm{g}}{208.98 \mathrm{g/mol}} \approx 0.0478 \mathrm{mol}\] 2. Determine the moles of electrons used in the process: \[n_{e} = 0.0478 \mathrm{mol}\] 3. Calculate the charge needed for the process: \[Q = 0.0478 \mathrm{mol} \times 96485 \mathrm{C/mol} \approx 4611.47 \mathrm{C}\] 4. Calculate the time required for electrolysis: \[t = \frac{4611.47 \mathrm{C}}{25.0 \mathrm{A}} \approx 184.46 \mathrm{s}\] So, it would take approximately 184.46 seconds to produce 10.0 grams of bismuth by the electrolysis of a BiO⁺ solution using a current of 25.0 A.