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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Numbers
Understanding oxidation numbers is key to identifying oxidation-reduction reactions. An oxidation number is a value assigned to an element in a chemical compound that represents the number of electrons lost or gained by an atom of that element during the formation of the compound. These numbers help track how electrons are distributed among atoms. Here鈥檚 how oxidation numbers are assigned to elements in various compounds:
  • For a pure element like diatomic oxygen ( O_2 ), the oxidation number is always 0 because the element is not combined with anything else.
  • In compounds, oxygen generally has an oxidation number of -2. However, in peroxides like hydrogen peroxide ( H_2O_2 ), oxygen has an oxidation number of -1.
  • Hydrogen typically has an oxidation number of +1 when combined with non-metals and -1 when combined with metals.
  • The sum of oxidation numbers in a neutral compound is zero, while in polyatomic ions, it equals the charge of the ion.
Using oxidation numbers, we can find out which substances are oxidized and which are reduced during the reaction. For example, manganese ( Mn ) in permanganate ( MnO_4^鈭� ) has an oxidation number of +7, which decreases to +4 in manganese dioxide ( MnO_2 ), indicating a reduction.
Electron Transfer
Electron transfer is the process where electrons move from one atom to another during a chemical reaction. This transfer is what constitutes oxidation-reduction or redox reactions. In such reactions, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Let鈥檚 see how electron transfer takes place using reaction (a) as an example:
For sulfur ( S ), the oxidation number goes from -2 in S^{2-} to 0 in S , implying that sulfur is oxidized. This indicates it loses electrons and its oxidation number increases. The loss of electrons (oxidation) by sulfur is matched by the gain of electrons (reduction) by manganese in MnO_4^鈭� , where the oxidation number of Mn changes from +7 to +4, signifying it is reduced. For each sulfur atom that loses 2 electrons, manganese receives 3 electrons. So when observing the stoichiometry, 6 electrons in total are transferred as multiple sulfur atoms interact with permanganate ions.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. During redox processes, one element gets oxidized, losing electrons, while another gets reduced, gaining electrons.
Here are key points to understand redox reactions:
  • Oxidation is the increase in oxidation state due to electron loss.
  • Reduction is the decrease in oxidation state through electron gain.
  • The substances taking part in oxidation and reduction are respectively called the reducing agent and the oxidizing agent. The reducing agent is oxidized, while the oxidizing agent is reduced.
Consider reaction (b) for example: In the reaction involving hydrogen peroxide ( H_2O_2 ) and dichlorine heptoxide ( Cl_2O_7 ), the oxidation number for oxygen in H_2O_2 increases from -1 to 0 when forming O_2 . Hence, H_2O_2 acts as a reducing agent, losing electrons in the process. Cl, going from +7 in Cl_2O_7 to +3 in ClO_2^鈭� , is reduced, acting as the oxidizing agent. This balanced change in oxidation number through electron transfer makes the reaction a redox process.

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Most popular questions from this chapter

A voltaic cell that uses the reaction $$ \mathrm{T}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tl}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) $$ has a measured standard cell potential of \(+1.19 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 M, P_{\mathrm{H}_{2}}=\) \(0.95 \mathrm{atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 M,\) and the \(\mathrm{pH}\) in both half-cells is 4.00\(?\)

A voltaic cell is constructed with two \(\mathrm{Zn}^{2+}-\) Zn electrodes. The two half-cells have \(\left[\mathrm{Zn}^{2+}\right]=1.8 M\) and \(\left[\mathrm{Zn}^{2+}\right]=1.00 \times 10^{-2} M,\) respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Zn}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

You may have heard that "antioxidants" are good for your health. Is an "antioxidant" an oxidizing agent or a reducing agent? [Sections 20.1 and 20.2\(]\)

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) , and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\) . The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.
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