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Chapter 19: Problem 2

Balance the following redox equations by the halfreaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution)

Short Answer

Expert verified
The balanced redox equations are: (a) \( \mathrm{2Mn^{2+} + 2H_2O_2 \rightarrow 2MnO_2 + 4H_2O} \), (b) \( \mathrm{2Bi(OH)_3 + 2SnO_2^{2-} \rightarrow 2SnO_3^{2-} + 2Bi + 3H_2O} \), (c) \( \mathrm{Cr_2O_7^{2-} + 6C_2O_4^{2-} \rightarrow 2Cr^{3+} + 6CO_2 + 14H^+} \), (d) \( \mathrm{ClO_3^{-} + 5Cl^- + 6H^+ \rightarrow 3Cl_2 + ClO_2 + 3H_2O} \).

Step by step solution

01

Identify the Oxidation and Reduction Half-Reactions for each equation

In every redox reaction, there's a substance that's being reduced and another one that's being oxidized. For the given equations, the substances being reduced or oxidized are: (a) \( \mathrm{Mn^{2+}} \) is being oxidized to \( \mathrm{MnO_2} \), (b) \( \mathrm{Bi(OH)_3} \) is being reduced to \( \mathrm{Bi} \), (c) \( \mathrm{Cr_2O_7^{2-}} \) is being reduced to \( \mathrm{Cr^{3+}} \), (d) \( \mathrm{ClO_3^{-}} \) is being reduced to \( \mathrm{ClO_2} \).
02

Balance each half-reaction for mass and charge

For each half-reaction, first balance the atoms other than \( \mathrm{H} \) and \( \mathrm{O} \), then add \( \mathrm{H_2O} \) to balance \( \mathrm{O} \) atoms, and if needed, add \( \mathrm{H^+} \) to balance \( \mathrm{H} \) atoms. Lastly, add electrons \( \mathrm{e^{-}} \) to balance the charge: (a) \( \mathrm{Mn^{2+} \rightarrow MnO_2 + 4H^+ + 2e^-} \), (b) \( \mathrm{Bi(OH)_3 + 3e^- \rightarrow Bi + 3OH^-} \), (c) \( \mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O} \), (d) \( \mathrm{ClO_3^{-} + 2H^+ + e^- \rightarrow ClO_2 + H_2O} \).
03

Balance the overall redox equations

Multiply the half-reactions by suitable factors so that the number of electrons in the two half-reactions are equal. Then, add the two half-reactions together and simplify to get the final balanced redox equations: (a) \( \mathrm{2Mn^{2+} + 2H_2O_2 \rightarrow 2MnO_2 + 4H_2O} \), (b) \( \mathrm{2Bi(OH)_3 + 2SnO_2^{2-} \rightarrow 2SnO_3^{2-} + 2Bi + 3H_2O} \), (c) \( \mathrm{Cr_2O_7^{2-} + 6C_2O_4^{2-} \rightarrow 2Cr^{3+} + 6CO_2 + 14H^+} \), (d) \( \mathrm{ClO_3^{-} + 5Cl^- + 6H^+ \rightarrow 3Cl_2 + ClO_2 + 3H_2O} \).

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Most popular questions from this chapter

How does the tendency of iron to rust depend on the \(\mathrm{pH}\) of solution?

The oxidation of \(25.0 \mathrm{~mL}\) of a solution containing \(\mathrm{Fe}^{2+}\) requires \(26.0 \mathrm{~mL}\) of \(0.0250 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in acidic solution. Balance the following equation and calculate the molar concentration of \(\mathrm{Fe}^{2+}\): $$ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{H}^{+} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+} $$

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