In: Chemistry
Which statement about the following redox reaction in acidic
solution is not correct?
MnO4??(aq) +
I???(aq) ? MnO2(s) +
I2(aq)
A) H2O and H?+ can be used to balance the oxygen atoms, if necessary, in each of the half-reactions. | |
B) MnO4??(aq) ? MnO2(s) is the reduction step. | |
C) I???(aq) ? I2(aq) is the oxidation step. | |
D) The redox atoms are already balanced in this reaction. |
+7 -1 +4 0
MnO4-(aq)
+ I-(aq)
MnO2(s) +
I2(aq)
Here the oxidation number of Mn decreases from +7 to +4 so it is reduced so it is the reduction step
& the reduction step is : MnO4-(aq)
MnO2(s)
Statement (B) is correct
The oxidation number of Iodine increases from -1 to 0 , it gets
oxidised, so it is the oxidation step & the oxidation step is :
I-(aq)
I2(aq)
Statement (C) is correct.
Oxygen atoms are balanced by adding the required no. of H2O molecules to the side deficient in Oxygen atoms & then balance Hydrogen atoms. For this add the required number of H+ ions to the side deficient in H atoms.
So option (A) is correct.
The reaction is not balalanced at all
So the incorrect statement is (D)
Therefore the correct option is (D)
NOTE :-
Rules for balancing of equation in ion electrode method:
(1) Identify the species that are oxidized and reduced.
(2) Divide the equation into oxidation half reaction & reduction half reaction
(3) Balance the atoms & charges in each half reaction separately according to the following rules:
(a) First balance the atoms other than that of Hydrogen & Oxygen
(b) Oxygen atoms are balanced by adding the required no. of H2O molecules to the side deficient in Oxygen atoms:
(c) Then balance Hydrogen atoms. For this
(i)If the reaction is in acidic medium add the required number of H+ ions to the side deficient in H atoms
(ii) If the reaction is in basic medium add the required number of H2O molecules to the side deficient in H atoms and then number of OH- ions to the other side.
(4) Add required number of electrons to balance the charges
(5) Equalize the number of electrons in the half reactions by multiplying each equation with the required number.
(6) Add the two reactions. so that cancel off the electrons takes place
Calculate the net number of common terms and write down on the proper side.