Question

Balance the following half-equations. Balance (a) and (b) in basic medium, \((\mathrm{c})\) and \((\mathrm{d})\) in acidic medium. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

Short answer

To balance the given half-equations, follow these steps: a) In basic medium: 1. Balance elements: Already balanced. 2. Balance oxygens: Add 2 H2O to the right side. 3. Balance hydrogens: Add 4 OH- to the left side. 4. Balance charges: Add 4e- to the right side. Final balanced equation: O2(g) + 4OH-(aq) → 2O2-(aq) + 2H2O(l) + 4e- b) In basic medium: 1. Balance elements: Already balanced. 2. Balance oxygens: Add 2 H2O to the right side. 3. Balance hydrogens: Add 4 OH- to the left side. 4. Balance charges: Add 3e- to the right side. Final balanced equation: MnO4-(aq) + 4OH-(aq) → MnO2(s) + 2H2O(l) + 3e- c) In acidic medium: 1. Balance elements: 2Cr on both sides. 2. Balance oxygens: Add 7 H2O to the right side. 3. Balance hydrogens: Add 14 H+ to the left side. 4. Balance charges: Add 6e- to the left side. Final balanced equation: Cr2O7^2-(aq) + 14H+(aq) + 6e- → 2Cr^3+(aq) + 7H2O(l) d) In acidic medium: 1. Balance elements: 2Cl on both sides. 2. Balance oxygens: No oxygens to balance. 3. Balance hydrogens: No hydrogens to balance. 4. Balance charges: Add 2e- to the right side. Final balanced equation: 2Cl^-(aq) → Cl2(g) + 2e-

Step-by-step solution

Balance elements other than O and H

As we have only oxygens, we move to Step 2.

Balance oxygens by adding H2O

To balance the oxygens, add 2 \(\mathrm{H}_{2}\mathrm{O}\) to the right-hand side of the equation: \(\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{O}^{2-}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\).

Balance hydrogens by adding OH-

We have 4 hydrogen atoms on the right-hand side, we add 4 \(\mathrm{OH}^{-}\) to the left-hand side: \(\mathrm{O}_{2}(g) + 4\mathrm{OH}^{-}(aq) \longrightarrow 2\mathrm{O}^{2-}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\).

Balance charges by adding electrons

Since there are 4 extra negative charges on the left-hand side due to \(\mathrm{OH}^{-}\) ions, add 4 electrons to the right-hand side to balance the charges: \(\mathrm{O}_{2}(g) + 4\mathrm{OH}^{-}(aq) \longrightarrow 2\mathrm{O}^{2-}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) + 4e^{-}\). (b) Balancing the half-equation in basic medium:

Balance elements other than O and H

Balance Manganese by ensuring there's just one Mn on each side: \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\).

Balance oxygens by adding H2O

To balance oxygens, add 2 \(\mathrm{H}_{2}\mathrm{O}\) to the right-hand side: \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l)\).

Balance hydrogens by adding OH-

We have 4 hydrogen atoms on the right-hand side, so add 4 \(\mathrm{OH}^{-}\) to the left-hand side: \(\mathrm{MnO}_{4}^{-}(a q) + 4\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{MnO}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l)\).

Balance charges by adding electrons

Since there are 3 extra negative charges on the left-hand side, add 3 electrons to the right-hand side to balance the charges: \(\mathrm{MnO}_{4}^{-}(a q) + 4\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{MnO}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) + 3e^{-}\). (c) Balancing the half-equation in acidic medium:

Balance elements other than O and H

In this case, we need to balance chromium. Since there are two Cr on the left-hand side, write 2Cr on the right-hand side: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q)\).

Balance oxygens by adding H2O

To balance the oxygens, add 7 \(\mathrm{H}_{2}\mathrm{O}\)s to the right-hand side: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\).

Balance hydrogens by adding H+

Since 14 hydrogen atoms are on the right side, add 14 \(\mathrm{H}^{+}\)s to the left side: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\).

Balance charges by adding electrons

Count the total charges on each side. On the left side, there are 6 positive charges and 2 negative charges. On the right side, there are 6 positive charges. To balance the charges, add 6 electrons to the left side: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q) + 6e^{-} \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\). (d) Balancing the half-equation in acidic medium:

Balance elements other than O and H

Balance chlorine by ensuring there are two Cl on each side: \(2\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\).

Balance oxygens by adding H2O

There are no oxygens to balance, so move to Step 3.

Balance hydrogens by adding H+

There are no hydrogens to balance, so move to Step 4.

Balance charges by adding electrons

Since there are 2 extra negative charges on the left-hand side due to \(\mathrm{Cl}^{-}\) ions, add 2 electrons to the right-hand side to balance the charges: \(2\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g) + 2e^{-}\).