Question

A potassium salt of an acid of molybdenum has the formula ${{K}_{2}}Mo{{O}_{x}}$. When as acidified solution of ${{K}_{2}}Mo{{O}_{x}}$ is electrolyzed between Pt electrodes, ${{O}_{2}}$ gas is liberated corresponding to a volume of 0.121 liters at STP and 0.3454 g of Mo is deposited. If the oxidation state of oxygen in the salt is -2, calculate the value of ‘x’. (Atomic mass of Mo is 96).

Answer 1

Hint: First, we have to write the reaction of electrolysis of water which will help to find the amount of current passed in the solution. This amount of current passed in the solution is equal to the moles of molybdenum produced in the solution. From this, we can calculate the oxidation state of molybdenum in the salt.

Complete answer:
Let us first write the reaction of electrolysis of water producing oxygen gas. The reaction will be:
$2{{H}_{2}}O+4{{e}^{-}}\to 4{{H}^{+}}+{{O}_{2}}$
We are given that the amount of oxygen gas liberated is 0.121 liters at STP and as we know that at STP the volume of any gas is 22.4 liters. So, the moles of ${{O}_{2}}$ liberated will be:
${{O}_{2}}\text{ moles = }\dfrac{0.121}{22.4}$
As we can see that there are 4 electrons involved in the reaction. So, the amount of current passed will be:
$Current=\dfrac{0.121}{22.4}\text{ x 4 F}$
We are given that the amount of molybdenum deposited in the reaction is 0.3454and the atomic mass of molybdenum is 96. The number of moles of Mo will be:
$Moles=\dfrac{0.3454}{96}$
Let Mo was in +n oxidation state then;
$M{{o}^{n+}}+n{{e}^{-}}\to Mo$
So, we can calculate the oxidation state of Mo by taking the ratio of current passed in the solution to the number of moles of Mo.
$Oxidation\text{ }state\text{ = }\dfrac{\dfrac{0.121}{22.4}\text{ x 4}}{\dfrac{0.3454}{96}}$
$Oxidation\text{ }state(n)=6$
So, the oxidation state of Mo is +6.
Given the formula of salt is ${{K}_{2}}Mo{{O}_{x}}$. Now, by putting the value of oxidation states, we can find the value of x.
$(+1)2+6+x(-2)=0$
$x=4$
So, the formula will be ${{K}_{2}}Mo{{O}_{4}}$.

Note:
Molybdenum is metal and mostly the oxidation state of the metal is positive, that's why we have taken the oxidation number of Mo as +n. Since the salt was neutral, the sum of oxidation states of all the elements is equal to zero.