Question

na2cro4 + h2o2 in acidic medium
What will be the color of the solution?

Answer 1

Step 1. Write down the unbalanced equation('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.

 

Cr(OH)4- + H2O2 → CrO42- + H2O

 

Step 2. Separate the redox reaction into half-reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

 

Cr+3(O-2H+1)4- + H+12O-12 → Cr+6O-242- + H+12O-2

 

b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). When one member of the redox couple is oxygen with an oxidation state of -2 or hydrogen with an oxidation state of +1, it is best to replace it with a water molecule.

O:

Cr+3(O-2H+1)4- → Cr+6O-242-

 

R:

H+12O-12 → H+12O

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas. Never change a formula when balancing an equation. Balance each half reaction separately.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

O:

Cr(OH)4- → CrO42-

 

R:

H2O2 → 2H2O

 

b) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:

Cr(OH)4- → CrO42-

 

R:

H2O2 → 2H2O

 

c) Balance the hydrogen atoms. Check if there are the same numbers of hydrogen atoms on the left and right side, if they aren't equilibrate these atoms by adding protons (H+).

O:

Cr(OH)4- → CrO42- + 4H+

 

R:

H2O2 + 2H+ → 2H2O

d) For reactions in a basic medium, add one OH- ion to each side for every H+ ion present in the equation. The OH- ions must be added to both sides of the equation to keep the charge and atoms balanced. Combine OH- ions and H+ ions that are present on the same side to form water.

O:

Cr(OH)4- + 4OH- → CrO42- + 4H2O

 

R:

H2O2 + 2H2O → 2H2O + 2OH-

 

Step 4. Balance the charge. To balance the charge, add electrons (e-) to the more positive side to equal the less positive side of the half-reaction. It doesn't matter what the charge is as long as it is the same on both sides.

O:

Cr(OH)4- + 4OH- → CrO42- + 4H2O + 3e-

 

R:

H2O2 + 2H2O + 2e- → 2H2O + 2OH-

Step 5. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:

Cr(OH)4- + 4OH- → CrO42- + 4H2O + 3e-

| *2

R:

H2O2 + 2H2O + 2e- → 2H2O + 2OH-

| *3

O:

2Cr(OH)4- + 8OH- → 2CrO42- + 8H2O + 6e-

 

R:

3H2O2 + 6H2O + 6e- → 6H2O + 6OH-

Step 6. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

2Cr(OH)4- + 3H2O2 + 8OH- + 6H2O + 6e- →2CrO42- + 14H2O + 6OH- + 6e-


Step 7. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

2Cr(OH)4- + 3H2O2 + 2OH- → 2CrO42- + 8H2O


Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type and number of atoms on both sides of the equation.


ELEMENTLEFTRIGHTDIFFERENCECr2*12*10O2*4 + 3*2 + 2*12*4 + 8*10H2*4 + 3*2 + 2*18*20

Second, verify that the sum of the charges on one side of the equation is equal to the sum of the charges on the other side. It doesn't matter what the charge is as long as it is the same on both sides.

2*-1 + 3*0 + 2*-1  =  2*-2 + 8*0


-4  =  -4

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, and since the charges on both sides are equal we can write a balanced equation.

2Cr(OH)4- + 3H2O2 + 2OH- →2CrO42- + 8H2O

Answer 2

Answer:

blue due to the formation of CrO5