Question

Purification of chromium can be achieved by electrorefining chromium from an impure chromium anode onto a pure chromium cathode in an electrolytic cell. How many hours will it take to plate 11.0 kg of chromium onto the cathode if the current passed through the cell is held constant at 41.5 A? Assume the chromium in the electrolytic solution is present as Cr₃. ______ h.

Answer 1

Explanation:

It is given that mass is 11 kg. Convert mass into grams as follows.

                $11 kg \times \frac{1000 g}{1 kg}$

               = 11000 g               (as 1 kg = 1000 g)

Now, calculate the number of moles as follows.

    No. of moles = $\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}$

                          = $\frac{11000 g}{52 g/mol}$

                          = 211.54 mol

For $Cr^{3+}$, 3 moles of electrons are required

Hence,       $3 \times 211.54 mol$

                  = 634.62 mol

As 1 mol of electrons contain 96500 C of charge. Therefore, charge carried by 634.62 mol of electrons will be calculated as follows.

             Q = $634.62 mol \times 96500 C/mol$

                  = $61.24 \times 10^{6} C$ of charge

We know that relation between charge, current and time is as follows.

                    Q = $I \times t$

Current is given as 41.5 A and charge is calculated as $61.24 \times 10^{6} C$. Therefore, calculate the time as follows.

                Q = $I \times t$

      $61.24 \times 10^{6} C = 41.5 A \times t$        

                   t = $1.47 \times 10^{6} sec$

As there are 3600 seconds in one 1 hour. Therefore, converting $1.47 \times 10^{6} sec$ into hours as follows.

            $\frac{1.47 \times 10^{6} sec}{3600 sec/hr}$

               = 4.09 hr

Thus, we can conclude that it takes 4.09 hours to plate 11.0 kg of chromium onto the cathode if the current passed through the cell is held constant at 41.5 A.