Question

A. Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga (s) that can be deposited from a Ga (III) solution by a current of 0.790 A that flows for 30.0 min?
B. A current of 5.79 A is passed through a Sn (NO3)2 solution. How long in hours would this current have to be applied to plate out 8.70 g of tin?

Answer 1

Answer:

A)Mass of  gallium plated out is 0.3440 grams

B) For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

Explanation:

To calculate the total charge, we use the equation:

$C=I\times t$

where,

C = Charge

I = Current in time t (seconds)

To calculate the moles of electrons, we use the equation:

$\text{Moles of electrons}=\frac{C}{F}$

where,

F = Faraday's constant = 96500

A) The equation for the deposition of Ga(s) from Ga(III) solution follows:

$Ga^{3+}(aq.)+3e^-\rightarrow Ga(s)$

I = 0.790 A, t = 30.0 min = 1800 seconds

$C=I\times t$

$C=0.790 A\times 1800 s=1422 C$

Moles of electron transferred:

$=\frac{1422 C}{96500 F}=0.01474 mol$

Now, to calculate the moles of gallium, we use the equation:

$\text{Moles of Gallium}=\frac{\text{Moles of electrons}}{n}$

n = number of electrons transferred = 3

$\text{Moles of Gallium}=\frac{0.01474 mol}{3}=0.004913 mol$

Mass of 0.004913 moles of gallium = 0.004913 mol × 70 g/mol=0.3440 g

B) The equation for the deposition of Sn(s) from Sn(II) solution follows:

$Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$

Moles of tin = $\frac{8.70 g}{119 g/mol}=0.07311 mol$

n = number of electrons transferred = 2

$\text{Moles of tin}=\frac{\text{Moles of electrons}}{n}$

Moles of electron =  $n\times \text{Moles of tin}$

$=2\times 0.07311 mol=0.14622 mol$

Charge transferred during time t :

$\text{Moles of electrons}=\frac{C}{F}$

$C=96500 F\times 0.14622 mol=14,110.23 C$

Current applied for t time = I = 5.79 A

$t=\frac{C}{I}=\frac{14,110.23 C}{5.79 A}=2,437 s=0.67 hrs$

For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.