Question

Question 40.5 pts

Redox reactions during Qualitative Analysis.

Likewise, there were several instances during the Qual procedure where a redox reaction occurred (which often was also associated with a change in solubility). During the cation procedure, adding a mixture of H₂O₂ and NaOH caused Cr(OH)₃ to be converted to CrO₄^2–. Not accidentally, one of these compounds was quite insoluble while the other one was quite soluble. The balanced reaction equation is:

2 Cr(OH)₃(s) + 3 H₂O₂(aq) + 4 OH^–(aq) → 2 CrO₄^2–(aq) + 8 H₂O(ℓ)

a. For this question, focus on the change in the oxidation state of chromium. What is the oxidation state of chromium in the reactant and product compounds?

Cr(OH)3	CrO42–
[ Select ]                       ["+8", "+1", "+7", "+4", "+6", "+5", "+3", "+2"]	[ Select ]                       ["+2", "+8", "+5", "+4", "+6", "+7", "+3", "+1"]

Answer 1

Solution:

The given reaction is:

2 Cr(OH)₃(s) + 3 H₂O₂(aq) + 4 OH^–(aq) → 2 CrO₄^2–(aq) + 8 H₂O(l)

In this reaction, Cr(OH)3 is converted to CrO4^​​​2- .

The Oxidation state of central metal is calculated by the charge of atoms attached to it. The sum of Oxidation numbers of all the atoms attached is equal to the charge on the whole compelx. By this, the Oxidation number of atom having unknown Oxidation number is calculated.

So, Formula for Calculation of Oxidation numbers is:

Oxidation number of known atoms + Oxidation number of unknown atom = Charge on Complex.

Or, Oxidation number of Unknown atom = Charge on Complex - Oxidation number of known atoms

We know,  

Oxygen has a charge -2

Hydrogen with non metals has charge = +1

Hydrogen with Metals has charge = -1

1) Calculatiom for Cr(OH)₃

Here, Cr is bonded with 3 OH atoms. Suppose, Cr has Oxidation number "x'. In OH, O is bonded with H. As, O has -2 charge and H has +1 charge ( because H is bonded with non metal Oxygen ). So, the charge on overall OH is -1.

As 3 OH are attached, so total charge is -3. Now, the total charge on the complex is 0. So, the charge on Cr is:

X + (-3) = 0

Or, X = +3

So, Oxidation state of Cr in Cr(OH)₃ is +3.

2) Calculatiom for CrO_{​​​​4}^​​​2-

Here, Cr is bonded with 4 Oxygen atoms and total charge on the complex is -2. Suppose, Cr has Oxidation number "X". We know, O atom has -2 charge. So, 4 Oxygen atoms will have -8 charge. The complex has -2 charge. So, the equation becomes:

X + (-8) = -2

Or, X = -2 +8

Or, X = +6

Hence, Oxidation state of Cr in CrO₄^2– is +6.