Question

Balance the following reaction using the ion electron method.
${N_2}{H_4} + Cl{O_3}^ - \to NO + C{l^ - }\left( {Basic Medium} \right)$

Answer 1

We know that with the Ion-electron technique (likewise called the half-response strategy), the redox condition is isolated into two half-conditions - one for oxidation and one for decrease. Every one of these half-responses is adjusted independently and afterward joined to give the fair redox condition.

Complete answer:
Stage 1: All reactants and products should be known. For a superior outcome compose the response in ionic structure.
${N_2}{H_4} + Cl{O_3}^ - \to NO + C{l^ - }$
Stage 2: Separate the redox response into half-responses. A redox response is only both oxidation and decrease responses occurring all the while.
a) Assign oxidation numbers for every iota in the condition. Oxidation number (likewise called oxidation state) is a proportion of the level of oxidation of a molecule in a substance
${N^{ - 2}}_2{H^ + }{^1_4}{\text{ }} \to {\text{ }}N + 2{O^{ - 2}}{\text{ }}$
$C{l^ + } + 5{O^{2 - }}_3 \to C{l^{ - 1}}$
b) Identify and work out all redox couples in response. Recognize which reactants are being oxidized (the oxidation number increments when it responds) and which are being diminished (the oxidation number goes down).
$O:{\text{ }}{N^{ - 2}}_2{H^ + }{^1_4}{\text{ }} \to {\text{ }}{N^{ + 2}}{O^{ - 2}}$
$R:{\text{ }}Cl + 5{O^{ - 2}}_3 \to {\text{ }}C{l^{ - 1}}$
Stage 3: Equilibrium the iotas in every half response. A compound condition should have a similar number of particles of every component on the two sides of the condition. Add suitable coefficients (stoichiometric coefficients) before the synthetic recipes. Never change a recipe when adjusting a condition. Equilibrium every half response independently.
a) Balance any remaining molecules with the exception of hydrogen and oxygen. We can utilize any of the species that show up in the skeleton conditions for this reason. Remember that reactants ought to be added distinctly to one side of the condition and products to one side.
$O:{\text{ }}{N_2}{H_4} \to {\text{ }}2NO$
$R:ClO{3^ - } \to {\text{ }}C{l^ - }$
b) Balance the oxygen particles. Check if there are similar quantities of oxygen iotas on the left and right side, on the off chance that they aren't equilibrating these particles by adding water atoms.
$O:{N_2}{H_4} + 2{H_2}O \to 2NO$
$R:ClO{3^ - } \to C{l^ - } + 3{H_2}O$
c) Balance the hydrogen molecules. Check if there are similar quantities of hydrogen molecules on the left and right side, on the off chance that they aren't equilibrated by adding protons.
$O:NH4 + 2H2O \to 2NO + 8{H^ + }$
$R:Cl{O^{3 - }} + 6{H^ + } \to C{l^ - } + 3{H_2}O$
d) For responses in an essential medium, add one $OH$ -particle to each side for each ${H^ + }$ ion present in the condition. The $OH$-particles should be added to the two sides of the condition to keep the charge and molecules adjusted. Consolidate $OH$-particles and ${H^ + }$particles that are available on a similar side to frame water.
$O:{N_2}{H_4} + 2{H_2}O + 8O{H^ - } \to 2NO + 8{H_2}O$
$R:ClO_3^1 + 6{H_2}O \to C{l^ - } + 3{H_2}O + 6O{H^ - }$
Stage 4: Equilibrium the charge. To adjust the charge, add electrons to the more certain side to rise to the more negative side of the half-response. It doesn't make any difference what the charge is, the length of it is something very similar on the two sides.
$O:{N_2}{H_4} + 2{H_2}O + 8O{H^ - } \to 2NO + 8{H_2}O + 8{e^ - }$
$R:ClO_3^ - + 6{H_2}O + 6{e^ - } \to C{l^ - } + 3{H_2}O + 6O{H^ - }$
Stage 5: Make electrons acquire identical to electrons lost. The electrons lost in the oxidation half-response should be equivalent to the electrons acquired in the decreased half-response. To make the two equivalent, increase the coefficients of all species by whole numbers delivering the least regular number between the half-responses.
$O:{N_2}{H_4} + 2{H_2}O + 8O{H^ - } \to 2NO + 8{H_2}O + 8{e^ - }$
$R:ClO_3^ - + 6{H_2}O + 6{e^ - } \to C{l^ - } + 3{H_2}O + 6O{H^ - }$
$O:3{N_2}{H_4} + 6{H_2}O + 24O{H^ - } \to 6NO + 24{H_2}O + 24{e^ - }$
$R:4ClO_3^ - + 24{H_2}O{\text{ }} + 24{e^ - } \to 4C{l^ - } + 12{H_2}O + 24OH$
Stage 6: Add the half-responses together. The two half-responses can be consolidated very much like two mathematical conditions, with the bolt filling in as the equivalents sign. Recombine the two half-responses by adding every one of the reactants together on one side and the entirety of the products together on the opposite side.
$3{N_2}{H_4} + 4ClO_3^ - + 30{H_2}O + 24O{H^ - } + 24{e^ - } \to 6NO + 4C{l^ - } + 36{H_2}O + 24{e^ - } + 24O{H^ - }$Stage 7: Improve on the condition. Similar species on the inverse sides of the bolt can be dropped. Compose the condition with the goal that the coefficients are the littlest arrangement of numbers conceivable.
$3{N_2}{H_4} + 4ClO_3^ - \to 6NO + 4C{l^ - } + 6{H_2}O$

Note:
We know that the Balance condition precisely depicts the amounts of reactants and items in compound responses. The Law of Conservation of Mass expresses that mass is neither made nor obliterated in a standard synthetic response. This implies that a compound condition should have a similar number of molecules of every component on both sides of the condition. Additionally the amount of the charges on one side of the condition should be equivalent to the amount of the charges on the opposite side. At the point when these two conditions are met, the condition is supposed to be adjusted.