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Maru [420]
3 years ago
13

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the a

mount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.590 A that flows for 30.0 min.
Chemistry
1 answer:
Sophie [7]3 years ago
6 0

Answer: 0.256 g

Explanation:

The deposition of Ga from Ga(III) solution

Ga^{3+}+3e^-\rightarrow Ga

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

I=\frac{q}{t}

where,

I = current passed = 0.590 A

q = total charge = ?

t = time required = 30.0 min = 1800 s (1min=60s)

Putting values in above equation, we get:

0.590A=\frac{q}{1800}\\\\q=1062C

Ga^{3+}+3e^-\rightarrow Ga

96500\times 3=289500C will deposit = 70 g of Gallium

1062C will deposit = \frac{70}{289500}\times 1062=0.256 g of Gallium

Thus 0.256 g of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.590 A that flows for 30.0 min.

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