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

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.

Step-by-step explanation:

To calculate the total charge, we use the equation:

`C=I* t`

where,

C = Charge

I = Current in time t (seconds)

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

`\text{Moles of electrons}=(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* t`

`C=0.790 A* 1800 s=1422 C`

Moles of electron transferred:

`=(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}=(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 =
`(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* \text{Moles of tin}`

`=2* 0.07311 mol=0.14622 mol`

Charge transferred during time t :

`\text{Moles of electrons}=(C)/(F)`

`C=96500 F* 0.14622 mol=14,110.23 C`

Current applied for t time = I = 5.79 A

`t=(C)/(I)=(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.