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Chapter 6: Problem 90

Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. \(\mathrm{Cl}_{2}(g)+\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathbf{O}_{2}(g)+\mathbf{H}_{2} \mathbf{O}(l)+\mathbf{P b}(s) \rightarrow \mathbf{P b}(\mathbf{O H})_{2}(s)\) c. \(\mathrm{H}^{+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(i)\)

Short Answer

Expert verified
The balanced equations for the given redox reactions are: a. \(3 Al(s) + 6 Cl_{2}(g) \rightarrow 3 Al^{3+}(aq) + 12 Cl^-(aq)\) b. \(2Pb(s) + O_2(g) + 2H_2O(l) \rightarrow 2Pb(OH)_2(s)\) c. \(8H^+(aq) + MnO_4^-(aq) + 5Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\)

Step by step solution

01

(a) Identify oxidation states

: We start by identifying the oxidation states of all the species involved: \(Cl_2\): \(Cl_{1}\) and \(Cl_{2}\) have oxidation states of 0 each. \(Al(s)\): The oxidation state of metallic Al is 0. \(Al^{3+}(aq)\): The oxidation state of Al is +3. \(Cl^-(aq)\): The oxidation state of Cl is -1.
02

(a) Determine change in oxidation states

: Now, we'll determine the change in oxidation states between reactants and products: \(Al\): Changes from an oxidation state of 0 to +3. \(Cl_{2}\): Both \(Cl_{1}\) and \(Cl_{2}\) change from an oxidation state of 0 to -1.
03

(a) Balance half-reactions

: Balance the half-reactions by balancing the atoms and adding electrons where appropriate: Oxidation half-reaction: \[Al \rightarrow Al^{3+} + 3e^-\] Reduction half-reaction: \[2 Cl_{2} + 6e^- \rightarrow 4 Cl^-\]
04

(a) Combine half-reactions

: Combine the balanced half-reactions. To do this, we need to multiply each half-reaction by the appropriate factor so they have the same number of electrons: 3(oxidation half-reaction) + 2(reduction half-reaction): \[3Al + 6 Cl_{2} \rightarrow 3 Al^{3+} + 12 Cl^-\] The balanced equation for reaction (a) is: \[3 Al(s) + 6 Cl_{2}(g) \rightarrow 3 Al^{3+}(aq) + 12 Cl^-(aq)\]
05

(b) Identify oxidation states

: First, we identify the oxidation states of the species involved in the reaction: \(O_2\): The oxidation state of O in \(O_2\) is 0. \(H_2 O\): The oxidation state of H is +1 and that of O is -2. \(Pb(s)\): The oxidation state of metallic Pb is 0. \(Pb(OH)_2\): The oxidation state of Pb is +2, O is -2 and H is +1.
06

(b) Determine change in oxidation states

: Next, we determine the change in oxidation states between reactants and products: \(O_2\): Changes from an oxidation state of 0 to -2. \(Pb(s)\): Changes from an oxidation state of 0 to +2.
07

(b) Balance half-reactions

: Balance the half-reactions by balancing the atoms and adding electrons where appropriate: Oxidation half-reaction: \[Pb \rightarrow Pb^{2+} + 2e^-\] Reduction half-reaction: \[O_2 + 2H_2O + 4e^- \rightarrow 4 OH^-\]
08

(b) Combine half-reactions

: Combine the balanced half-reactions: 2(oxidation half-reaction) + 1(reduction half-reaction): \[2Pb + O_2 + 2H_2O \rightarrow 2Pb^{2+} + 4 OH^-\] The balanced equation for reaction (b) is: \[2Pb(s) + O_2(g) + 2H_2O(l) \rightarrow 2Pb(OH)_2(s)\]
09

(c) Identify oxidation states

: Identify the oxidation states of species involved in the reaction: \(H^+\): The oxidation state of H is +1. \(MnO_4^-\): The oxidation state of Mn is +7, O is -2. \(Fe^{2+}\): The oxidation state of Fe is +2. \(Mn^{2+}\): The oxidation state of Mn is +2. \(Fe^{3+}\): The oxidation state of Fe is +3. \(H_2O\): The oxidation state of H is +1 and that of O is -2.
10

(c) Determine change in oxidation states

: Determine the change in oxidation states between reactants and products: \(MnO_4^-\): Mn changes from an oxidation state of +7 to +2. \(Fe^{2+}\): Changes from an oxidation state of +2 to +3.
11

(c) Balance half-reactions

: Balance the half-reactions by balancing the atoms and adding electrons where appropriate: Oxidation half-reaction: \[5 Fe^{2+} \rightarrow 5 Fe^{3+} + 5e^-\] Reduction half-reaction: \[MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\]
12

(c) Combine half-reactions

: Combine the balanced half-reactions: 1(oxidation half-reaction) + 1(reduction half-reaction): \[MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\] The balanced equation for reaction (c) is: \[8H^+(aq) + MnO_4^-(aq) + 5Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\]

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Most popular questions from this chapter

A solution was prepared by mixing \(50.00 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HNO_and \(100.00 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HNO \(_{3}\). Calculate the molarity of the final solution of nitric acid.

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, I ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0 \mathrm{ppm}\) is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{mL},\) which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). A 1.45 -g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of \(\mathrm{Al}(\mathrm{OH})_{3}\). The precipitate is filtered, dried, and weighed. The mass of the precipitate is \(0.107 \mathrm{g} .\) What is the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample?

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?

A 450.0 -mL sample of a 0.257 - \(M\) solution of silver nitrate is mixed with \(400.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?
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