Question

Balance the below equation using the half-reaction method.
$Ag(s)+C{{N}^{-}}(aq)+{{O}_{2}}(g)\to Ag(CN)_{2}^{-}(aq)$

Answer 1

In the half-reaction method, the redox equation is separated into two half-equations, one for oxidation and one for reduction. Balance each of these half-reactions separately and then combine to give the balanced redox equation.

Complete step by step answer:
- We will balance the given equation, using the ion- exchange, also known as half-reaction method.
- Firstly, write the unbalanced equation of the chemical reaction
$Ag(s)+C{{N}^{-}}(aq)+{{O}_{2}}(g)\to Ag(CN)_{2}^{-}(aq)+O{{H}^{-}}(aq)$
- Since, a redox reaction is both oxidation and reduction reactions taking place simultaneously, we will separate the process into two half reactions
After assigning oxidation number for each atom in the equation we get, $\overset{0}{\mathop{Ag}}\,+\overset{+2-3}{\mathop{C{{N}^{-}}}}\,+\overset{0}{\mathop{{{O}_{2}}}}\,\overset{{}}\leftrightarrows\overset{+1}{\mathop{Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+\overset{-2+1}{\mathop{OH}}\,$
- Now, we will identify which reactants are being oxidized and which are being reduced and note the transfer of electrons.
Oxidation- $\overset{0}{\mathop{Ag}}\,\to \overset{+1}{\mathop{Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+{{e}^{-}}$
Reduction- $\overset{{}}{\mathop{{{\overset{0}{\mathop{O}}\,}_{2}}+4{{e}^{-}}}}\,\to \overset{+1-2}{\mathop{2{{H}_{2}}O}}\,$
- Now, we will balance the atoms in each half reaction
Oxidation- $\overset{0}{\mathop{Ag}}\,+\overset{{}}{\mathop{2C{{N}^{-}}}}\,\to \overset{+1}{\mathop{Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+{{e}^{-}}$
Reduction- $\overset{{}}{\mathop{{{\overset{0}{\mathop{O}}\,}_{2}}+4{{e}^{-}}}}\,\to \overset{+1-2}{\mathop{2{{H}_{2}}O}}\,$
- After balancing the charge in both equations, we get,
Oxidation- $\overset{0}{\mathop{Ag}}\,+\overset{{}}{\mathop{2C{{N}^{-}}}}\,\to \overset{+1}{\mathop{Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+{{e}^{-}}$
Reduction- $\overset{{}}{\mathop{{{\overset{0}{\mathop{O}}\,}_{2}}+4{{e}^{-}}}}\,\to \overset{+1-2}{\mathop{2{{H}_{2}}O}}\,+4O{{H}^{-}}$
- Now, balance the oxygen atoms to attain same number of atoms both sides
Oxidation- $\overset{0}{\mathop{Ag}}\,+\overset{{}}{\mathop{2C{{N}^{-}}}}\,\to \overset{+1}{\mathop{Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+{{e}^{-}}$
Reduction- $\overset{{}}{\mathop{{{\overset{0}{\mathop{O}}\,}_{2}}+4{{e}^{-}}}}\,+4{{H}_{2}}O\to \overset{+1-2}{\mathop{2{{H}_{2}}O}}\,+4O{{H}^{-}}$
- Now we make electron gain equivalent to electron loss by making the two reactions equal. Multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.
Oxidation- $\overset{0}{\mathop{4Ag}}\,+\overset{{}}{\mathop{8C{{N}^{-}}}}\,\to \overset{+1}{\mathop{Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+4{{e}^{-}}$
Reduction- $\overset{{}}{\mathop{{{\overset{0}{\mathop{O}}\,}_{2}}+4{{e}^{-}}}}\,+4{{H}_{2}}O\to \overset{+1-2}{\mathop{2{{H}_{2}}O}}\,+4O{{H}^{-}}$
- The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals’ sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side. After adding these two reactions we get,
$\overset{0}{\mathop{4Ag}}\,+\overset{{}}{\mathop{{{\overset{0}{\mathop{O}}\,}_{2}}\overset{{}}{\mathop{+8C{{N}^{-}}}}\,+4{{e}^{-}}+4{{H}_{2}}O}}\,\overset{{}}\leftrightarrows\overset{+1}{\mathop{4Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+2\overset{+1-2}{\mathop{{{H}_{2}}O}}\,+4{{e}^{-}}+4O{{H}^{-}}$
- Now, after simplify the equation the equation we get,
$\overset{0}{\mathop{4Ag}}\,+\overset{{}}{\mathop{{{\overset{0}{\mathop{O}}\,}_{2}}\overset{{}}{\mathop{+8C{{N}^{-}}}}\,+2{{H}_{2}}O}}\,\overset{{}}\leftrightarrows\overset{+1}{\mathop{4Ag}}\,(\overset{+2-3}{\mathop{CN}}\,)_{2}^{-}+4O{{H}^{-}}$
Therefore, our final balanced equation is

& \overset{{}}{\mathop{4Ag}}\,+\overset{{}}{\mathop{{{\overset{{}}{\mathop{O}}\,}_{2}}\overset{{}}{\mathop{+8C{{N}^{-}}}}\,+2{{H}_{2}}O}}\,\overset{{}}\leftrightarrows\overset{{}}{\mathop{4Ag}}\,(\overset{{}}{\mathop{CN}}\,)_{2}^{-}+4O{{H}^{-}} \\\ & \\\ \end{aligned}$$ **Note:** The presence of hydroxide ions means that the reaction takes place in acidic solution. Usually, most redox reactions will actually only proceed in one type of solution or the other.