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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:

User Cleong
by
8.3k points

1 Answer

4 votes

Step-by-step 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} * 96500}

molar mass of Al = 26.98 g/mol

Z =
(26.98 g/mol)/(3 * 96500)

=
9.32 * 10^(-5) g/mol

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

W = Zit

=
9.32 g/mol * 10^(-5) * 0.834 amp * 1.175 * 3600 sec

= 0.328 g

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

User Levi Rosol
by
7.0k points
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