# Balance the following equations in both acidic and basic environments. Show the half-reactions...

## Question:

Balance the following equations in both acidic and basic environments. Show the half-reactions right before they are added together (after balancing all the components) and then show the balanced equation.

a. {eq}Cr_2O_7^{2-}(aq) + C_2H_5OH(l) \to Cr^{3+}(aq) + CO_2(g) {/eq}

b. {eq}Fe^{2+}(aq) + MnO_4^-(aq) \to Fe^{3+}(aq) + Mn^{2+}(aq) {/eq}

## Balancing Redox Reaction:

To balance the redox reaction, we will divide the reaction into oxidation and reduction half cell reactions and individually balanced them. We will use hydrogen ions in balancing the redox reaction in acidic medium whereas in basic medium, hydroxide ions are used which are added in balanced redox reaction in acidic medium so that all hydrogen ions are neutralized by hydroxide ions.

#### Part a:

{eq}\rm Cr_2O_7^{2-}(aq) + C_2H_5OH(l) \to Cr^{3+}(aq) + CO_2(g) {/eq}

To balance the above redox reaction, we will divide the reaction as oxidation and reduction reaction:

{eq}\rm Cr_2O_7^{2-}(aq) \to Cr^{3+}(aq) \\ C_2H_5OH(l) \to CO_2(g) {/eq}

We will balance in acidic medium firstly, that is hydrogen ions are used.

Balance all the elements other than H and O firstly.

{eq}\rm Cr_2O_7^{2-}(aq) \to 2Cr^{3+}(aq) \\ C_2H_5OH(l) \to 2CO_2(g) {/eq}

Now, balance the O atoms and for this we will add water molecules.

{eq}\rm Cr_2O_7^{2-}(aq) \to 2Cr^{3+}(aq) + 7H_2O \\ C_2H_5OH(l) + 3H_2O \to 2CO_2(g) {/eq}

Now, balance the H atoms and this we will add the hydrogen ions in acidic medium.

{eq}\rm Cr_2O_7^{2-}(aq) + 14H^+ \to 2Cr^{3+}(aq) + 7H_2O \\ C_2H_5OH(l) + 3H_2O \to 2CO_2(g) + 12H^+ {/eq}

Now, we will balanced the charge by adding electrons.

{eq}\rm Cr_2O_7^{2-}(aq) + 14H^+ + 6e^- \to 2Cr^{3+}(aq) + 7H_2O \\ C_2H_5OH(l) + 3H_2O \to 2CO_2(g) + 12H^+ + 12e^- {/eq}

Now, we will balanced the electrons:

{eq}\rm 2Cr_2O_7^{2-}(aq) + 28H^+ + 12e^- \to 4Cr^{3+}(aq) + 14H_2O \\ C_2H_5OH(l) + 3H_2O \to 2CO_2(g) + 12H^+ + 12e^- {/eq}

Now, add the above 2 equations to get the balanced redox reaction in acidic medium:

{eq}\rm \boxed{\rm 2Cr_2O_7^{2-}(aq) + 16H^+ + C_2H_5OH(l) \to 4Cr^{3+}(aq) + 11H_2O + 2CO_2(g)} {/eq}

To balanced the reaction in basic medium., we will add 16 moles of hydroxide ions on the both sides of the above equation to neutralize the hydrogen ions and form water molecules.

{eq}\rm 2Cr_2O_7^{2-}(aq) + 16H^+ + 16OH^- + C_2H_5OH(l) \to 4Cr^{3+}(aq) + 11H_2O + 2CO_2(g) + 16OH^- {/eq}

Hydrogen and hydroxide ions combine to form water molecules.

{eq}\rm \boxed{\rm 2Cr_2O_7^{2-}(aq) + 16H_2O + C_2H_5OH(l) \to 4Cr^{3+}(aq) + 11H_2O + 2CO_2(g) + 16OH^-} {/eq}

#### Part b:

{eq}\rm Fe^{2+}(aq) + MnO_4^-(aq) \to Fe^{3+}(aq) + Mn^{2+}(aq) {/eq}

To balance the above redox reaction, we will divide the reaction as oxidation and reduction reaction:

{eq}\rm MnO_4^-(aq) \to Mn^{2+}(aq) \\ Fe^{2+}(aq) \to Fe^{3+}(aq) {/eq}

We will balance in acidic medium firstly, that is hydrogen ions are used.

Now, balance the O atoms and for this we will add water molecules.

{eq}\rm MnO_4^-(aq) \to Mn^{2+}(aq) + 4H_2O \\ Fe^{2+}(aq) \to Fe^{3+}(aq) {/eq}

Now, balance the H atoms and this we will add the hydrogen ions in acidic medium.

{eq}\rm MnO_4^-(aq) + 8H^+ \to Mn^{2+}(aq) + 4H_2O \\ Fe^{2+}(aq) \to Fe^{3+}(aq) {/eq}

Now, we will balanced the charge by adding electrons.

{eq}\rm MnO_4^-(aq) + 8H^+ + 5e^- \to Mn^{2+}(aq) + 4H_2O \\ Fe^{2+}(aq) \to Fe^{3+}(aq) + e^- {/eq}

Now, we will balanced the electrons:

{eq}\rm MnO_4^-(aq) + 8H^+ + 5e^- \to Mn^{2+}(aq) + 4H_2O \\ 5Fe^{2+}(aq) \to 5Fe^{3+}(aq) + 5e^- {/eq}

Now, add the above 2 equations to get the balanced redox reaction in acidic medium:

{eq}\rm \boxed{\rm MnO_4^-(aq) + 8H^+ + 5Fe^{2+}(aq) \to Mn^{2+}(aq) + 4H_2O + 5Fe^{3+}(aq) } {/eq}

To balanced the reaction in basic medium., we will add 16 moles of hydroxide ions on the both sides of the above equation to neutralize the hydrogen ions and form water molecules.

{eq}\rm MnO_4^-(aq) + 8H^+ + 8OH^- + 5Fe^{2+}(aq) \to Mn^{2+}(aq) + 4H_2O + 5Fe^{3+}(aq) + 8OH^- {/eq}

Hydrogen and hydroxide ions combine to form water molecules.

{eq}\rm \boxed{\rm MnO_4^-(aq) + 8H_2O + 5Fe^{2+}(aq) \to Mn^{2+}(aq) + 4H_2O + 5Fe^{3+}(aq) + 8OH^-} {/eq}