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Chapter 4: Problem 94

Balance the following oxidation–reduction reactions that occur in basic solution. a. $\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)$ b. $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)$ c. $\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)$

Short Answer

Expert verified
The balanced oxidation-reduction reactions in basic solution are: a. \(Cr(s) + CrO_{4}^{2-}(aq) + 8 H_2O \rightarrow 2\ Cr(OH)_{3}(s) + 6 OH^-(aq)\) b. \(2 MnO_4^-(aq) + S_2^{2-}(aq) + 16 H_2O \rightarrow 2 MnS(s) + S(s) + 16 OH^-(aq)\) c. \(3 CN^-(aq) + MnO_4^-(aq) + 2 H_2O \rightarrow 3 CNO^-(aq) + MnO_2(s) + 4 OH^-(aq)\)

Step by step solution

01

Identify the half-reactions

: First, we need to recognize the oxidation and reduction half-reactions. Oxidation half-reaction: \(\mathrm{Cr}(s) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) Reduction half-reaction: \(\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\)
02

Balance the half-reactions

: Next, we need to balance each half-reaction individually. Oxidation half-reaction: \(Cr \rightarrow Cr(OH)_3 \(+ 3 e^-) \\ The half-reaction is already balanced. Reduction half-reaction: \(CrO_4^{2-} + 8 H_2O \rightarrow Cr(OH)_3 + 6 OH^-(aq) + 3 e^-\)
03

Combine the balanced half-reactions

: Lastly, we need to sum the balanced half-reactions to form the balanced overall reaction: \(\mathrm{Cr}(s) + \mathrm{CrO}_{4}^{2-}(a q) + 8 H_2O \rightarrow 2\ \mathrm{Cr}(\mathrm{OH})_{3}(s) + 6 OH^-(aq)\) b. Balance the reaction: MnO4^-(aq) + S2-(aq) -> MnS(s) + S(s)
04

Identify the half-reactions

: Oxidation half-reaction: \(\mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnS}(s)\)
05

Balance the half-reactions

: Oxidation half-reaction: \(S^{2-} \rightarrow S(s) + 2 e^-\) Reduction half-reaction: \(MnO_4^- + 8 H_2O + 5 e^- \rightarrow MnS + 8 OH^-\)
06

Combine the balanced half-reactions

: 2 MnO4^-(aq) + S2-(aq) + 16 H2O -> 2 MnS(s) + S(s) + 16 OH^-(aq) c. Balance the reaction: CN^-(aq) + MnO4^-(aq) -> CNO^-(aq) + MnO2(s)
07

Identify the half-reactions

: Oxidation half-reaction: \(\mathrm{CN}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)\)
08

Balance the half-reactions

: Oxidation half-reaction: \(CN^- \rightarrow CNO^- + e^-\) Reduction half-reaction: \(MnO_4^- + 2 H_2O + 3 e^- \rightarrow MnO_2 + 4 OH^-\)
09

Combine the balanced half-reactions

: 3 CN^-(aq) + MnO4^-(aq) + 2 H2O -> 3 CNO^-(aq) + MnO2(s) + 4 OH^-(aq)

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Most popular questions from this chapter

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

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