Question

A metal forms the fluoride MF3. Electrolysis of the molten fluoride by a current of 3.86 A for 16.2 minutes deposits 1.25 g of the metal. Calculate the molar mass of the metal.

Answer 1

Answer: The molar mass of the metal is 96.45 g/mol

Explanation:

The fluoride of the metal formed is $MF_3$

The oxidation half-reaction follows:

$M\rightarrow M^{3+}+3e^-$

Calculating the theoretical mass deposited by using Faraday's law, which is:

$m=\frac{M\times I\times t(s)}{n\times F}$       ......(1)

where,

m = actual mass deposited = 1.25 g

M = molar mass of metal = ?

I = average current = 3.86 A

t = time period in seconds = 16.2 min = 972 s            (Conversion factor: 1 min = 60 sec)

n = number of electrons exchanged = $3mol^{-1}$

F = Faraday's constant = 96500 C

Putting values in equation 1, we get:

$1.25g=\frac{M\times 3.86A\times 972s}{3mol^{-1}\times 96500 C}\\\\M=\frac{1.25g\times 3mol^{-1}\times 96500 C}{3.86A\times 972s}\\\\M=96.45g/mol$

Hence, the molar mass of the metal is 96.45 g/mol