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Chapter 20: Problem 18

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{S}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{S}(s)+} \\ {\quad 2 \mathrm{MnO}_{2}(s)+8 \mathrm{OH}^{-}(a q)} \\ {\text { (b) } 4 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(g)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+} \\ {\quad 5 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{O}_{2}(g)} \\\\{\text { (c) } \mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \longrightarrow} \\ {\quad \mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)}\end{array} $$

Short Answer

Expert verified
In the given redox reactions: (a) 6 electrons are transferred, with S(-II) -> S(0) having a change in oxidation number of +2, and Mn(+VII) -> Mn(+IV) having a change of -3. (b) 4 electrons are transferred, with O(-I) in \(H_2O_2\) -> O(0) in \(O_2\) having a change in oxidation number of +1, and Cl(+VII) -> Cl(+III) having a change of -4. (c) 1 electron is transferred, with O(-I) in \(H_2O_2\) -> O(0) in \(O_2\) having a change in oxidation number of +1, and no electrons transferred for Cl(+IV) -> Cl(+IV).

Step by step solution

01

(a) Assign Oxidation Numbers

For reaction (a), we assign the oxidation numbers for each element in both reactants and products: \(MnO_4^-\): Mn has an oxidation number of +7, while O has -2. \(S^{2-}\): S has an oxidation number of -2. \(H_2O\): H has an oxidation number of +1, while O has -2. \(S(s)\): S has an oxidation number of 0. \(MnO_2\): Mn has an oxidation number of +4, while O has -2. \(OH^-\): O has an oxidation number of -2, and H has +1.
02

(a) Calculate Electron Transfer

For reaction (a), we determine the change in oxidation numbers for the species involved in redox reactions: For S(-II) -> S(0), the change in oxidation number is +2. For Mn(+VII) -> Mn(+IV), the change in oxidation number is -3. Now, multiply the change in oxidation numbers by the stoichiometric coefficients to determine the total number of electrons transferred: The total number of electrons gained by S is \(3 \times 2 = 6\). The total number of electrons lost by Mn is \(2 \times 3 = 6\). In reaction (a), 6 electrons are transferred.
03

(b) Assign Oxidation Numbers

For reaction (b), we assign the oxidation numbers for each element: \(H_2O_2\): H has an oxidation number of +1, while O has -1. \(Cl_2O_7\): Cl has an oxidation number of +7, while O has -2. \(OH^-\): O has an oxidation number of -2, and H has +1. \(ClO_2^-\): Cl has an oxidation number of +3, while O has -2. \(H_2O\): H has an oxidation number of +1, while O has -2. \(O_2\): O has an oxidation number of 0.
04

(b) Calculate Electron Transfer

For reaction (b), we determine the change in oxidation numbers for the species involved in redox reactions: For O(-I) in \(H_2O_2\) -> O(0) in \(O_2\), the change in oxidation number is +1. For Cl(+VII) -> Cl(+III), the change in oxidation number is -4. Multiply the change in oxidation numbers by the stoichiometric coefficients to determine the total number of electrons transferred: The total number of electrons gained by O is \(4 \times 1 = 4\). The total number of electrons lost by Cl is \(1 \times 4 = 4\). In reaction (b), 4 electrons are transferred.
05

(c) Assign Oxidation Numbers

For reaction (c), we assign the oxidation numbers: \(Ba^{2+}\): Ba has an oxidation number of +2. \(OH^-\): O has an oxidation number of -2, and H has +1. \(H_2O_2\): H has an oxidation number of +1, while O has -1. \(ClO_2\): Cl has an oxidation number of +4, while O has -2. \(Ba(ClO_2)_2\): Ba has an oxidation number of +2, Cl has +4 and O has -2. \(H_2O\): H has an oxidation number of +1, while O has -2. \(O_2\): O has an oxidation number of 0.
06

(c) Calculate Electron Transfer

For reaction (c), we determine the change in oxidation numbers for the species involved in redox reactions: For O(-I) in \(H_2O_2\) -> O(0) in \(O_2\), the change in oxidation number is +1. For Cl(+IV) -> Cl(+IV) (no change), no electrons are transferred. Multiply the change in oxidation numbers by the stoichiometric coefficients to determine the total number of electrons transferred: The total number of electrons gained by O is \(1 \times 1 = 1\). In reaction (c), 1 electron is transferred.

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Most popular questions from this chapter

If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

Gold exists in two common positive oxidation states, \(+1\) and \(+3 .\) The standard reduction potentials for these oxidation states are $$ \begin{array}{ll}{\mathrm{Au}^{+}(a q)+\mathrm{e}^{-}} \quad {\longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.69 \mathrm{V}} \\\ {\mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)} \quad {E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{V}}\end{array} $$ (a) Can you use these data to explain why gold does not tarnish in the air? ( b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{array}{rl}{4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+2} & {\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)} \\ {\longrightarrow} & {4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q)}\end{array} $$ What is being oxidized, and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode half-cell is 1.00\(M\) and the cell generates an emf of \(+0.22 \mathrm{V},\) what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? (b) If the anode half-cell contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

A common shorthand way to represent a voltaic cell is $$ \text {anode} | \text {anode solution} | | \text {cathode solution} | \text {cathode} $$ A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fel Fe \(^{2+} \| \operatorname{Ag}^{+} | A g;\) calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by Zn \(\left|Z \mathrm{n}^{2+}\right| \mathrm{H}^{+} | \mathrm{H}_{2} ;\) calculate the standard cell emf using data in Appendix E and use Pt for the hydrogen electrode. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \\ \longrightarrow & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the ClO \(_{3}^{-}\) and \(\mathrm{Cl}^{-} .\) Calculate the standard cell emf given: \(\mathrm{ClO}_{3}^{-}(a q)+\) \(6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l); E^{\circ}=1.45 \mathrm{V}\).

Given the following half-reactions and associated standard reduction potentials: $$ \begin{array}{c}{\text { AuBr }_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{V}} \\ {\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.
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