Question

Write the stoichiometric coefficient for the following reaction:
$x{I_2} + yO{H^ - }\xrightarrow{{}}I{O_3}^ - + z{I^ - } + 3{H_2}O$
A.$x = 6,y = 3,z = 5$
B.$x = 3,y = 2,z = 3$
C.$x = 3,y = 6,z = 5$
D.$x = 3,y = 3,z = 3$

Answer 1

A redox reaction involves the simultaneous occurrence of both oxidation and reduction processes. In oxidation, electrons are given out; whereas, in reduction, there is a gain of electrons. Now as we know that ultimately all processes are electrically neutral so the number of electrons lost is equal to the number of electrons gained.

Complete step by step answer:
Step 1
A redox reaction is balanced only when the electrons given out in the oxidation process are completely utilized in the reduction process. The two methods used for balancing oxidation-reduction reactions are:
Oxidation number method
Ion-electron method
Step 2
The given reaction has to be balanced in the ion-electron method in the basic medium.
In the ion-exchange method, the given reaction is split up into two half-reactions, one corresponding to the process of oxidation and the other to the process of reduction. The two half-reactions are balanced separately and then added in such a way that the electrons released in the process of oxidation cancel the electrons captured in the process of reduction.
Step 3:
Write the given reaction:
$x{I_2} + yO{H^ - }\xrightarrow{{}}I{O_3}^ - + z{I^ - } + 3{H_2}O$
Step 4
Write the oxidation numbers of all the atoms involved in the equation and find the species undergoing oxidation and reduction.
0 -2 +1 +5 -2 -1 +1 -2
$x{I_2} + yO{H^ - }\xrightarrow{{}}I{O_3}^ - + z{I^ - } + 3{H_2}O$
Oxidation ${I^0} - 5{e^ - }\xrightarrow{{}}{I^{ + 5}}$
Reduction ${I^0} + 1{e^ - }\xrightarrow{{}}{I^ - }$
Step 5
You need to balance both the equation to cancel the electrons lost or gained:
$({I^0} - 5{e^ - }\xrightarrow{{}}{I^{ + 5}}) \times 1$
$({I^0} + 1{e^ - }\xrightarrow{{}}{I^ - }) \times 5$
Step 6
In this step you need to balance all the other atoms other than hydrogen and oxygen:
$I + 5I + yO{H^ - }\xrightarrow{{}}I{O_3}^ - + 5{I^ - } + 3{H_2}O$
Step 7
Balance oxygen by adding a sufficient number of $O{H^ - }$ions on the side deficient in oxygen.
$I + 5I + 3O{H^ - }\xrightarrow{{}}I{O_3}^ - + 5{I^ - } + 3{H_2}O$
Step 8
For each hydrogen atom, you need to add a water molecule to the side deficient of the hydrogen atom. So, you have seen that there are 3 water molecules added to the product side. Now add an equal number of $O{H^ - }$ions to the opposite side.
$I + 5I + 3O{H^ - } + 3O{H^ - }\xrightarrow{{}}I{O_3}^ - + 5{I^ - } + 3{H_2}O$
Step 9
The final balanced reaction is:
$3{I_2} + 6O{H^ - }\xrightarrow{{}}I{O_3}^ - + 5{I^ - } + 3{H_2}O$
Hence the correct answer to the given question option c) $x = 3,y = 6,z = 5$.

So, the correct answer is “Option A”.

Note: In the ion-exchange method balancing in neutral and acidic solution also takes place where the process involves the adjustment through water and ${H^ + }$ions.