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Question
X––KMnO4+Y––K2C2O4+Z––H2SO4→A––Mn04+B––C02+6K2SO4+S––H2O
X,Y,Z,A,B and S are stoichiometric coefficients.
Here X,Y,Z, A and B are respectively
X,Y,Z,A,B and S are stoichiometric coefficients.
Here X,Y,Z, A and B are respectively
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Solution
The correct option is A 2, 5, 8, 2, 10
Let’s go step by step here.
Ion electron method: First write the given equation in ionic form having ions with central atom (which has undergone a change in oxidation state).
MnO−4+C2–––O2−4→Mn––––2++CO2
Note that O and H atoms attached to the central atom (shown in underlined) have to be retained.
Now, write the oxidation and reduction half reaction and balance them as shown:
Reduction:
MnO−4+C2–––O2−4→Mn––––2+
a. First, make sure that the element undergone the change in oxidation state is balanced.
b. Balance O atoms by adding 4H2O on RHS.
c. Now, RHS has excess of 8H atoms. Add 8H+ on LHS. Note, the medium is acidic due to the presence of H2SO4.
d. Now O and H are balanced. Balance the charge on both sides.
On LHS: Charge is 1×(−1)+8×(+1)=+7
On RHS: Charge is 1×(+2)+4×(0)=+2
Add 5e− in LHS (Note: each e− is equivalent to a charge of -1)
MnO−4+8H++5e−→Mn2++4H2O−−−−−−−−−Eq.(i)
Oxidation:
C2O2−4→CO2
Following the same procedure as above, we have:
a. Balance C atoms: C2O2−4→2CO2
b. Balance O atoms: Already balanced
c. Balance H atoms: No H atom
d. C2O2−4→2CO2+2e−−−−−−−−−−−−Eq(ii)
e. Multiply Eq. (i) by 2 and Eq. (ii) by 5 to balance the electrons transfer and add to get
2MnO−4+5C2O2−4+16H+→2Mn2++10CO2+8H2O
f. Now add other ions to both sides
L.H.S(12k+)R.H.S.(12K+) 8SO2−48SO2−4
g. 2K++2MnO−4+10K++5C2O2−4+8SO2−4+8SO2−4+16H+ ↓ 2Mn2++2SO2−4+10CO2+12K++6SO2−4+8H2O2KMnO4+5K2C2O4+8H2SO4 ↓2MnSO4+10CO2+6K2SO4+8H2O
Let’s go step by step here.
Ion electron method: First write the given equation in ionic form having ions with central atom (which has undergone a change in oxidation state).
MnO−4+C2–––O2−4→Mn––––2++CO2
Note that O and H atoms attached to the central atom (shown in underlined) have to be retained.
Now, write the oxidation and reduction half reaction and balance them as shown:
Reduction:
MnO−4+C2–––O2−4→Mn––––2+
a. First, make sure that the element undergone the change in oxidation state is balanced.
b. Balance O atoms by adding 4H2O on RHS.
c. Now, RHS has excess of 8H atoms. Add 8H+ on LHS. Note, the medium is acidic due to the presence of H2SO4.
d. Now O and H are balanced. Balance the charge on both sides.
On LHS: Charge is 1×(−1)+8×(+1)=+7
On RHS: Charge is 1×(+2)+4×(0)=+2
Add 5e− in LHS (Note: each e− is equivalent to a charge of -1)
MnO−4+8H++5e−→Mn2++4H2O−−−−−−−−−Eq.(i)
Oxidation:
C2O2−4→CO2
Following the same procedure as above, we have:
a. Balance C atoms: C2O2−4→2CO2
b. Balance O atoms: Already balanced
c. Balance H atoms: No H atom
d. C2O2−4→2CO2+2e−−−−−−−−−−−−Eq(ii)
e. Multiply Eq. (i) by 2 and Eq. (ii) by 5 to balance the electrons transfer and add to get
2MnO−4+5C2O2−4+16H+→2Mn2++10CO2+8H2O
f. Now add other ions to both sides
L.H.S(12k+)R.H.S.(12K+) 8SO2−48SO2−4
g. 2K++2MnO−4+10K++5C2O2−4+8SO2−4+8SO2−4+16H+ ↓ 2Mn2++2SO2−4+10CO2+12K++6SO2−4+8H2O2KMnO4+5K2C2O4+8H2SO4 ↓2MnSO4+10CO2+6K2SO4+8H2O
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