Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate the amount of Ga ( s ) that can be deposited from a Ga ( III ) solution using a current of 0.220 A that flows for 40.0 min .

Amount of Gallium = 0.127g

Explanation:

Electrolysis equation is:

Ga3+   +   3e-    > Ga

 To calculate the charge

t = 40.0 min = 40.0 x 60 s = 2400 s

time, t = 2400s

Q = I*t =  

= 0.22A x 2400s

= 528 C

1 mol of Ga requires 3 mol of electron

1 mol of electron = 1 Faraday =96485 C

So,1 mol of Ga requires  96485x 3= 289455 C

mol of Gallium = 528/289455 = 0.00182 mol

Molar mass of Ga = 69.72 g/mol

mass of Ga = number of moles x  molar mass

= 0.00182mol * 69.72

g/mol

= 0.127g

or you can use this direct formula

m=(current*time/Faraday's)*(molar mass/no of electrons transferred)

keeping in mind   Ga3+ + 3e- → Ga

n=3

m=(It/F)*(mew/n)

m =(0.22 x 2400/96485) x (69.72/3)

m=0.127 g

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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