In this article, we are going to learn the chemical reaction of Cr2o3 + Naoh + H2o , its product formation and balancing the chemical equation.
Before learning how the reaction Cr2o3 + Naoh + H2o proceeds we need to learn what kind of reagents are being used and what products would be formed depending on their conditions.
Cr2O3 , also known as Dichromium Trioxide is an inorganic oxide of Chromium in (+III) oxidation state that can be found in a rare mineral known as eskolaite, this oxide is generally green in colour which do not hold serious health hazards but can cause irritation to skin and lungs if inhaled for a longer period of time. Its molecular weight is 151.99 g/mol with a density of 5.21 g/cm3 .
NaOH , known by chemical name as Sodium hydroxide and commonly as caustic soda is an inorganic base and highly corrosive that can degrade a protein molecule in ambient conditions. It is easily soluble in water, absorbs moisture like CO2 and cause severe skin burns and allergy. Its molecular formula is 40 g/mol with density 2.13 g/cm3 .
H2O is the chemical formula for water, also referred to as dihydrogen monoxide with a molecular weight of 18 g/mol and a boiling point of 2120 F at 760 mm Hg. It is colorless, odourless, tasteless and most important solvent for living beings survival.
What is Cr2o3 + Naoh + H2o ?
Cr2o3 + Naoh + H2o is a chemical reaction between dichromium trioxide , sodium hydroxide and water that can be either a complexation or a redox reaction depending on the number of moles of alkali/base that reacts producing different complex products.
This reaction is mainly affected by the alkali’s concentration.
Cr element can undergo several types of reaction including redox, displacement, complexation, substitution reactions due to its varying oxidation states from +2 to +6 .
When 2 only moles of NaOH of is used, we get tetrahydroxochromate(II) complex, and when 6 moles of is used, we get hexahydroxochromate(III) complex . Both of which can act as reducing agents.
What is the product of Cr2o3 + Naoh + H2o ?
Cr2o3 + Naoh + H2o reacts to form a complex compound of Cr in +2 or +3 oxidation states with Na+ ion as the counterpart cationic species balancing the complex ion.
Depending on the number of moles of alkali/base that reacts, different complex molecules are formed with Cr. What actually varies in the chemical formula of different species is the hydroxy groups present in the complex ion that is affected by the concentration of alkali reacting in the reaction.
The first reaction is a redox reaction- an oxidation reduction reaction :
Cr2O3 + 2NaOH + 3H2O ——-> 2Na[Cr(OH)4]
- Sodium tetrahydroxochromate(II) is the product.
Cr is in +3 oxidation state in reactant form and in +2 oxidation state in product form.
The second reaction is a complexation reaction :
Cr2O3 + 6NaOH + 3H2O ——-> 2Na3[Cr(OH)6]
- Sodium hexahydroxochromate(III) is the product.
Cr is in +3 oxidation state in both reactant and product forms.
How to balance Cr2o3 + Naoh + H2o ?
There are two reactions here to be discussed.
The simplest way to balance a chemical equation is by first writing down the coefficient i.e. , the number that is written before the reactant as “X” , “Y” and so on , which denotes the number of moles that is used to carry out the reaction to completion.
The subscript in chemical formula represents the number of each constituent element that participates to form a molecule. An element containing subscript ‘1’ means it is present as an atom.
- yCr2O3 + xNaOH + zH2O ——-> v{Na[Cr(OH)4}
2. nCr2O3 + mNaOH + pH2O ——-> q{Na3[Cr(OH)6}
Step 1 : Count the number of atoms of each constituent element in both sides of the reaction and form a table.
- Cr = 2 , O = 5 , Na = 1 , H = 3 on the left side of the equation 1.
- Cr = 1 , O = 4 , Na = 1 , H = 4 on the right side of the equation 1.
- Cr = 2 , O = 5 , Na = 1 , H = 3 on the left side of the equation 2.
- Cr = 1 , O = 4 , Na = 1 , H = 6 on the right side of the equation 2.
Step 2 : Balance the elements which in this case is Cr and Na apart from H and O .
It is easier to balance O and H as these are common elements and they usually gets balanced while balancing other elements.
There is 2 Cr on the left side, and 1 on the right side, therefore, multiply right side by 2 as its coefficient , which is same for both the reactions ( a and b )
This makes Na unbalanced leaving 1 Na on left side and 2 Na atoms on the right side. So, multiply NaOH by 2 as its coefficient in case of a.
But a problem arises in b. as Na gets unbalanced by multiplying NaOH with 2 as there arises 6 Na atoms on the right side and only 2 on the left side. Therefore, multiply NaOH by 6 in b.
yCr2O3 + 2NaOH + zH2O ——-> 2{Na[Cr(OH)4} in case of a.
nCr2O3 + 6NaOH + pH2O ——-> 2{Na3[Cr(OH)6} in case of b.
This makes Cr and Na balanced on both sides.
Step 3 : Balance Hydrogen followed by balancing Oxygen.
There are 8 H on right side and 4 H on the left side of the equation. Multiplying H2O by the coefficient magnitude of 3. On doing so, we see that O is also balanced with a total of 8 oxygen atoms on both sides of the equation in a.
In b. , there are 12 H on the right side and 7 H on left side so multiplying H2O by 3. This gives 12 H atoms on both sides and O atoms get balanced in the meantime.
Balanced equation : a. Cr2O3 + 2NaOH + 3H2O ——-> 2Na[Cr(OH)4] ( less alkaline solution of NaOH )
b. Cr2O3 + 6NaOH + 3H2O ——-> 2Na3[Cr(OH)6] ( strong concentration of NaOH )
What type of reaction is Cr2o3 + Naoh + H2o ?
Cr2o3 + Naoh + H2o reaction type varies depending on the amount of alkali used. With less concentrated alkaline solution, Na[Cr(OH)4] is produced and in higher concentrated alkaline solution, Na3[Cr(OH)6] is produced.
Both a. and b. reaction produce complexes but the first one is a redox reaction as Cr undergoes a change in oxidation state from +3 to +2 forming a complex and in b. , Cr oxidation state doesn’t change forming a complex meaning it is a complexation reaction.
Conclusion :
Cr2o3 + Naoh + H2o reaction produces complexes that act as reducing agents in analytical chemistry, redox reactions, and forming other important complexes of Cr by substituting the hydroxy group, mainly applying in chelation chemistry.
Hello…. I am Nandita Biswas. I have completed my master’s in Chemistry with a specialization in organic and physical chemistry. Also, I have done two projects in chemistry- One dealing with colorimetric estimation and determination of ions in solutions. Others in Solvatochromism study fluorophores and their uses in the field of chemistry alongside their stacking properties on emission. I am working as a Research Associate Trainee in Medicinal Department.
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