Question

What mass (in g) of aluminum can be electroplated when 0.834 amps are used for 1.175 hours using a solution of aluminum nitrate? Answer:

Answer 1

Explanation:

The given chemical reaction will be as follows.

      $Al(NO_{3})_{3} \rightarrow Al(s) + 3e^{-}$

Also, mass deposited can be calculated using the formula as follows.

               W = Zit

where,     W = weight or mass of the substance

                Z = electrochemical equivalent

                 i = current

                 t = time in seconds

Calculate value of Z for the given reaction as follows.

               Z = $\frac{\text{molar mass of Al}}{\text{no. of electrons} \times 96500}$  

             molar mass of Al = 26.98 g/mol

                 Z = $\frac{26.98 g/mol}{3 \times 96500}$  

                    = $9.32 \times 10^{-5}$ g/mol

Therefore, putting the given values into the above formula as follows.

                    W = Zit

                         = $9.32 g/mol \times 10^{-5} \times 0.834 amp \times 1.175 \times 3600 sec$

                          = 0.328 g

Thus, we ca conclude that 0.328 g of aluminium can be electroplated in the given situation.