Question

How do you balance \[CO(g) + {I_2}{O_5}\left( s \right) \to {I_2}\left( s \right) + C{O_2}\left( g \right)\]?

Answer 1

Hint: A balanced chemical equation, the total number of atoms of each element in the reactants, on the left-hand side of the equation is the same as the number atoms in the products framed on the right-hand side of the equation
The complete mass of the reactants is equivalent to the absolute mass of the products.
The number of atoms of each element, before the reaction and after the reaction is equivalent.

Complete step by step answer:
You're really managing a redox reaction used to convert over carbon monoxide, \[CO\], to carbon dioxide, \[C{O_2}\] , by utilizing what is known as the Schutze reagent, which is iodine pentoxide,\[\;{I_2}{O_5}\] , on silica gel.
Iodine pentoxide goes about as an oxidizing agent here and oxidizes carbon monoxide to carbon dioxide while being reduced to iodine simultaneously.
Assign oxidation numbers to the atoms that partake in the reaction - for effortlessness, I won't add the conditions of the chemical species associated with the reaction
\[CO + {I_2}{O_5} \to {I_2} + C{O_2}\]
You can balance this condition by expecting that the reaction happens in acidic conditions (you will get the same result in the event that you accept essential conditions).
In this way, carbon's oxidation state goes from \[ + 2\] on the reactants' side to \[ + 4\] on the items' side, which obviously implies that it's being oxidized.
The oxidation half-condition will resemble this
\[CO{\text{ }} \to C{O_2} + 2{e^ - }\]
Here every carbon atom loses two electrons. Balance the oxygen atoms by adding water particles to the side that needs oxygen, and the hydrogen atoms by adding protons, \[{H^ + }\], to the side that needs hydrogen
\[{H_2}O + CO \to C{O_2} + 2{e^ - } + 2{H^ + }\]
The oxidation condition of iodine goes from \[ + 5\] on the reactant's side, to \[0\] on the items' side, which implies that it's being reduced.
The reduction half-reaction will be
\[{I_2}{O_5} + 10{e^ - } \to {I_2}\]
Here every iodine molecule gains five electrons, so two atoms will acquire a sum of ten electrons. Indeed, balance the oxygen and hydrogen to get
\[10{H^ + } + {I_2}{O_5} + 10{e^ - } \to {I_2} + 5{H_2}O\]
As you most likely are aware, in any redox reaction, the number of electrons lost in the oxidation half-condition should be equivalent to the number of electrons acquired in the reduction half-condition.
To get these two balanced, multiply the oxidation half-condition by \[5\]. Add the two half-conditions to get
\[\left[ {{H_2}O + CO + C{O_2} + 2{e^ - } + 2{H^ + }} \right] \times 5\]
\[10{H^ + } + {I_2}{O_5} + 10{e^ - } \to {I_2} + 5{H_2}O\]

The balanced chemical condition for this reaction will consequently be - state images included!
\[5CO\left( g \right) + {I_2}{O_5}\left( s \right) \to {I_2}\left( s \right) + 5C{o_2}\left( g \right)\]

Note: Use coefficients of products and reactants to adjust the number of atoms of an element on the two sides of a chemical equation.
At the point when an equivalent number of atoms of an element is available on the both sides of a chemical equation, the equation is balanced.