(a) Balance the following reaction by ion-electron method:. K3[Fe(CN)6] + N2H4 + KOH → N2 + K4[Fe(CN)6] + H2O
Answers
Answer:
Step 1. Write down the unbalanced equation
N2H4 + Fe(CN)63- + OH- → N2 + Fe(CN)64-
Step 2. Separate the Redox Reaction into half-reactions.
A Redox reaction is nothing, but both oxidation and reduction reactions taking place at the same time.
i) Assign oxidation numbers for each atom in the equation.
Oxidation number also known as oxidation state, that is a measure of the degree of oxidation, of an atom in a substance.
N-22H+14 + Fe+3(C+2N-3)63- + O-2H+1- → N02 + Fe+2(C+2N-3)64-
ii) Identify and Noted Down Every Redox Couples in Reaction.
O: N-22H+14 → N02 (N)
R: Fe+3(C+2N-3)63- → Fe+2(C+2N-3)64- (Fe)
iii) Combine these Redox Couples into two half-reactions:
one for the oxidation,
and one for the reduction.
O: N-22H+14 → N02
R: Fe+3(C+2N-3)63- → Fe+2(C+2N-3)64-
Step 3. Balance the atoms in each half reaction.
i) Balance all other atoms other than hydrogen and oxygen.
O: N2H4 → N2
R: Fe(CN)63- → Fe(CN)64-
ii) Balance the oxygen atoms.
O: N2H4 → N2
R: Fe(CN)63- → Fe(CN)64-
iii) Balance the hydrogen atoms.
O: N2H4 → N2 + 4H+
R: Fe(CN)63- → Fe(CN)64-
iv) For reactions in a basic medium, add one OH- ion to each side for every H+ ion present in the equation.
Keep in mind to add OH- ions to both sides of the equation to keep the charge and atoms balanced. Then Combine OH- ions and H+ ions, which are present on the same side to form water.
O: N2H4 + 4OH- → N2 + 4H2O
R: Fe(CN)63- → Fe(CN)64-
Step 4. Balance the Charge.
O: N2H4 + 4OH- → N2 + 4H2O + 4e-
R: Fe(CN)63- + e- → Fe(CN)64-
Step 5. Make electron gain equivalent to electron lost.
O: N2H4 + 4OH- → N2 + 4H2O + 4e- | *1
R: Fe(CN)63- + e- → Fe(CN)64- | *4
O: N2H4 + 4OH- → N2 + 4H2O + 4e-
R: 4Fe(CN)63- + 4e- → 4Fe(CN)64-
Step 6. Add the half-reactions together.
N2H4 + 4Fe(CN)63- + 4OH- + 4e- → N2 + 4Fe(CN)64- + 4H2O + 4e-
Step 7. Simplify the equation.
N2H4 + 4Fe(CN)63- + 4OH- → N2 + 4Fe(CN)64- + 4H2O
Finally, always check that the equation is balanced.
First of all, verify that the equation contains the same type & number of atoms at the both sides of the equation.
Then, 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 same on both sides.
1*0 + 4*-3 + 4*-1 = 1*0 + 4*-4 + 4*0
-16 = -16
When you confirm that 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, it means whe have balanced the equation as below:
N2H4 + 4Fe(CN)63- + 4OH- → N2 + 4Fe(CN)64- + 4H2O