### Balancing Redox Equations

The unbalanced equation below describes the decomposition of molten sodium chloride:

$$NaCl(l)⟶Na(l)+Cl_2(g)\qquad unbalanced$$

This reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes *reduction*) and that for Cl is increased from −1 to 0 (it undergoes *oxidation*). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na:

$$2NaCl(l)⟶2Na(l)+Cl_2(g)\qquad balanced$$

Redox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, H^{+}(*aq*) and OH^{−}(*aq*), as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the *half-reaction method* is helpful. This approach involves the following steps:

- Write skeletal equations for the oxidation and reduction half-reactions.
- Balance each half-reaction for all elements except H and O.
- Balance each half-reaction for O by adding H
_{2}O. - Balance each half-reaction for H by adding H
^{+}. - Balance each half-reaction for charge by adding electrons.
- If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
- Add the two half-reactions and simplify.
- If the reaction takes place in a basic medium, add OH
^{−}ions the equation obtained in step 7 to neutralize the H^{+}ions (add in equal numbers to both sides of the equation) and simplify.

The examples below demonstrate the application of this method to balancing equations for aqueous redox reactions.

Balancing Equations for Redox Reactions in Acidic Solutions

Write the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.

Solution

Following the steps of the half-reaction method:

*Write skeletal equations for the oxidation and reduction half-reactions. $$\text{oxidation}\qquad Cu(s)⟶Cu^{2+}(aq)+Cu^{2+}(aq)$$ $$\text{reduction}\qquad HNO_3(aq)⟶NO(g)$$**Balance each half-reaction for all elements except H and O. $$\text{oxidation}\qquad Cu(s)⟶Cu^{2+}(aq)+Cu^{2+}(aq)$$ $$\text{reduction}\qquad HNO_3(aq)⟶NO(g)$$**Balance each half-reaction for O by adding H*_{2}O. $$\text{oxidation}\qquad Cu(s)⟶Cu^{2+}(aq)+Cu^{2+}(aq)$$ $$\text{reduction}\qquad HNO_3(aq)⟶NO(g)+2H_2O(l)$$*Balance each half-reaction for H by adding H*^{+}. $$\text{oxidation}\qquad Cu(s)⟶Cu^{2+}(aq)+Cu^{2+}(aq)$$ $$\text{reduction}\qquad 3H^+(aq)+HNO_3(aq)⟶NO(g)+2H_2O(l)$$*Balance each half-reaction for charge by adding electrons. $$\text{oxidation}\qquad Cu(s)⟶Cu^{2+}(aq)+2e^-$$ $$\text{reduction}\qquad 3e^-+3H^+(aq)+HNO_3(aq)⟶NO(g)+2H_2O(l)$$**If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other. $$\text{oxidation }(\times 3)\qquad 3Cu(s)⟶3Cu^{2+}(aq)+6\require{enclose}\enclose{horizontalstrike}{2}e^-$$ $$\text{reduction }(\times 2)\qquad 6\enclose{horizontalstrike}{3}e^-+6\enclose{horizontalstrike}{3}H^+(aq)+2HNO_3(aq)⟶2NO(g)+4\enclose{horizontalstrike}{2}H_2O(l)$$**Add the two half-reactions and simplify. $$3Cu(s)+\enclose{horizontalstrike}{6}e^-+6H^+(aq)+2HNO_3(aq)⟶3Cu^{2+}(aq)+\enclose{horizontalstrike}{6}e^-+2NO(g)+4H_2O(l)$$ $$3Cu(s)+6H^+(aq)+2HNO_3(aq)⟶3Cu^{2+}(aq)+2NO(g)+4H_2O(l)$$**If the reaction takes place in a basic medium, add OH*^{−}ions the equation obtained in step 7 to neutralize the H^{+}ions (add in equal numbers to both sides of the equation) and simplify.

This step not necessary since the solution is stipulated to be acidic.

The balanced equation for the reaction in an acidic solution is then

$$\mathbf{3Cu(s)+6H^+(aq)+2HNO_3(aq)⟶3Cu^{2+}(aq)+2NO(g)+4H_2O(l)}$$Check Your Learning

The reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction.

## Answer:

$$Cu(s)+2H^+(aq)+2HNO_3(aq)⟶Cu^{2+}(aq)+2NO_2(g)+2H_2O(l)$$

Balancing Equations for Redox Reactions in Basic Solutions

Write the balanced equation representing reaction between aqueous permanganate ion, $MnO_4^-$, and solid chromium(III) hydroxide, Cr(OH)_{3}, to yield solid manganese(IV) oxide, MnO_{2}, and aqueous chromate ion, $CrO_4^{2-}$. The reaction takes place in a basic solution.

Solution

Following the steps of the half-reaction method:

*Write skeletal equations for the oxidation and reduction half-reactions. $$\text{oxidation}\qquad Cr(OH)_3(s)⟶CrO_4^{2-}(aq)$$ $$\text{reduction}\qquad MnO_4^-(aq)⟶MnO_2(s)$$**Balance each half-reaction for all elements except H and O. $$\text{oxidation}\qquad Cr(OH)_3(s)⟶CrO_4^{2-}(aq)$$ $$\text{reduction}\qquad MnO_4^-(aq)⟶MnO_2(s)$$**Balance each half-reaction for O by adding H*_{2}O. $$\text{oxidation}\qquad H_2O(l)+Cr(OH)_3(s)⟶CrO_4^{2-}(aq)$$ $$\text{reduction}\qquad MnO_4^-(aq)⟶MnO_2(s)+2H_2O(l)$$*Balance each half-reaction for H by adding H*^{+}. $$\text{oxidation}\qquad H_2O(l)+Cr(OH)_3(s)⟶CrO_4^{2-}(aq)+5H^+(aq)$$ $$\text{reduction}\qquad 4H^+(aq)+MnO_4^-(aq)⟶MnO_2(s)+2H_2O(l)$$*Balance each half-reaction for charge by adding electrons. $$\text{oxidation}\qquad H_2O(l)+Cr(OH)_3(s)⟶CrO_4^{2-}(aq)+5H^+(aq)+3e^-$$ $$\text{reduction}\qquad 3e^-+4H^+(aq)+MnO_4^-(aq)⟶MnO_2(s)+2H_2O(l)$$**If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.*

This step is not necessary since the number of electrons is already in balance.*Add the two half-reactions and simplify. $$\enclose{horizontalstrike}{3e^-}+\enclose{horizontalstrike}{4H^+(aq)}+MnO_4^-(aq)+\enclose{horizontalstrike}{H_2O(l)}+Cr(OH)_3(s)⟶CrO_4^{2-}(aq)+\enclose{horizontalstrike}{5}H^+(aq)+\enclose{horizontalstrike}{3e^-}+MnO_2(s)+2H_2O(l)$$ $$MnO_4^-(aq)+Cr(OH)_3(s)⟶CrO_4^{2-}(aq)+H^+(aq)+MnO_2(s)+H_2O(l)$$**If the reaction takes place in a basic medium, add OH*$$OH^-(aq)+MnO_4^-(aq)+Cr(OH)_3(s)⟶CrO_4^{2-}(aq)+H^+(aq)+OH^-(aq)+MnO_2(s)+H_2O(l)$$ $$\mathbf{OH^-(aq)+MnO_4^-(aq)+Cr(OH)_3(s)⟶CrO_4^{2-}(aq)+MnO_2(s)+2H_2O(l)}$$^{−}ions the equation obtained in step 7 to neutralize the H^{+}ions (add in equal numbers to both sides of the equation) and simplify.

Check Your Learning

Aqueous permanganate ion may also be reduced using aqueous bromide ion, Br^{−}, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, BrO_{3}^{−}. Write the balanced equation for this reaction occurring in a basic medium.

## Answer:

$$H_2O(l)+2MnO_4^-(aq)+Br^-(s)⟶2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$$