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Chapter 18: Problem 32

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
a. 2Cr(s) + 6OH^-(aq) + 3CrO4^2-(aq) + 12H2O(l) -> 5Cr(OH)3(s) b. 5S^2-(aq) + 2OH^-(aq) + 2MnO4^-(aq) -> 5S(s) + 2Mn^2+(aq) + 2OH^-(aq) + H2O(l) + S(aq) c. 3CN^-(aq) + MnO4^-(aq) + 2H2O(l) -> 3CNO^-(aq) + MnO2(s) + 4OH^-(aq)

Step by step solution

01

Assign oxidation states

: Cr(s): 0 Cr in CrO4^2-: +6 O: -2 Cr in Cr(OH)3: +3 O in OH-: -2
02

Define half-reactions

: Oxidation: Cr -> Cr^3+ + 3e- Reduction: CrO4^2- + 6e- -> Cr^3+
03

Balance electrons

: To balance the electrons, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3: Oxidation: 2Cr -> 2Cr^3+ + 6e- Reduction: 3CrO4^2- + 18e- -> 3Cr^3+
04

Balance the charges in a basic solution

: Now we need to add hydroxide ions (OH-) to balance the charges for the basic solution: Oxidation: 2Cr + 6OH- -> 2Cr(OH)3 + 6e- Reduction: 3CrO4^2- + 18e- + 12H2O -> 3Cr(OH)3
05

Add half-reactions

: Now we can add the half-reactions to get the balanced equation: 2Cr + 6OH- + 3CrO4^2- + 12H2O -> 2Cr(OH)3 + 3Cr(OH)3 Simplifying, we get: 2Cr(s) + 6OH^-(aq) + 3CrO4^2-(aq) + 12H2O(l) -> 5Cr(OH)3(s) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\)
06

Assign oxidation states

: Mn in MnO4-: +7 O: -2 S^2-: -2 Mn in MnS: +2 S in MnS: -2 S: 0
07

Define half-reactions

: Oxidation: S^2- -> S + 2e- Reduction: MnO4^- + 5e- -> Mn^2+
08

Balance electrons

: To balance the electrons, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2: Oxidation: 5S^2- -> 5S + 10e- Reduction: 2MnO4^- + 10e- -> 2Mn^2+
09

Balance the charges in a basic solution

: Now we need to add hydroxide ions (OH-) to balance the charges for the basic solution: Oxidation: 5S^2- + 10OH- -> 5S + 5H2O + 10e- Reduction: 2MnO4^- + 10e- + 4H2O -> 2Mn^2+ + 8OH^-
10

Add half-reactions

: Now we can add the half-reactions to get the balanced equation: 5S^2-(aq) + 10OH^-(aq) + 2MnO4^-(aq) + 4H2O(l) -> 5S(s) + 5H2O(l) + 2Mn^2+(aq) + 8OH^-(aq) Simplifying, we get: 5S^2-(aq) + 2OH^-(aq) + 2MnO4^-(aq) -> 5S(s) + 2Mn^2+(aq) + 2OH^-(aq) + H2O(l) Finally, add S(aq) to get the final balanced equation: 5S^2-(aq) + 2OH^-(aq) + 2MnO4^-(aq) -> 5S(s) + 2Mn^2+(aq) + 2OH^-(aq) + H2O(l) + S(aq) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)
11

Assign oxidation states

: C in CN-: +2 N in CN-: -3 Mn in MnO4-: +7 O in MnO4-: -2 C in CNO-: +3 N in CNO-: -1 Mn in MnO2: +4 O in MnO2: -2
12

Define half-reactions

: Oxidation: CN^- -> CNO^- + e- Reduction: MnO4^- + 3e- -> MnO2
13

Balance electrons

: To balance the electrons, we need to multiply the oxidation half-reaction by 3: Oxidation: 3CN^- -> 3CNO^- + 3e- Reduction: MnO4^- + 3e- -> MnO2
14

Balance the charges in a basic solution

: Now we need to add the hydroxide ions (OH-) to balance the charges for the basic solution: Oxidation: 3CN^- -> 3CNO^- + 3e- Reduction: MnO4^- + 3e- + 2H2O -> MnO2 + 4OH^-
15

Add half-reactions

: Now we can add the half-reactions to get the balanced equation: 3CN^-(aq) + MnO4^-(aq) + 2H2O(l) -> 3CNO^-(aq) + MnO2(s) + 4OH^-(aq) This is the final balanced equation for the third redox reaction.

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Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{ccll}}^{\circ}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} M\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

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