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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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Most popular questions from this chapter

In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. [The Monitor and the CSS Virginia (formerly the USS Merrimack) fought the first battle between iron-armored ships.] In 1987 investigations were begun to see if the ship could be salvaged. It was reported in Time (June 22,1987 ) that scientists were considering adding sacrificial anodes of zinc to the rapidly corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.

The black silver sulfide discoloration of silverware can be removed by heating the silver article in a sodium carbonate solution in an aluminum pan. The reaction is $$3 \mathrm{Ag}_{2} \mathrm{~S}(s)+2 \mathrm{Al}(s) \rightleftharpoons 6 \mathrm{Ag}(s)+3 \mathrm{~S}^{2-}(a q)+2 \mathrm{Al}^{3+}(a q)$$ a. Using data in Appendix 4 , calculate \(\Delta G^{\circ}, K\), and \(\mathscr{C}^{\circ}\) for the above reaction at \(25^{\circ} \mathrm{C}\). (For \(\mathrm{Al}^{3+}(a q), \Delta G_{\mathrm{f}}^{\circ}=-480 . \mathrm{kJ} / \mathrm{mol}\).) b. Calculate the value of the standard reduction potential for the following half-reaction: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{~S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

Which of the following is the best reducing agent: \(\mathrm{F}_{2}, \mathrm{H}_{2}, \mathrm{Na}\), \(\mathrm{Na}^{+}, \mathrm{F}^{-}\) ? Explain. Order as many of these species as possible from the best to the worst oxidizing agent. Why can't you order all of them? From Table \(18.1\) choose the species that is the best oxidizing agent. Choose the best reducing agent. Explain.

. When jump-starting a car with a dead battery, the ground jumper should be attached to a remote part of the engine block. Why?
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