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Chapter 20: Problem 25

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\) HCOOH \((a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s)+\mathrm{OCl}^{-}(a q) \longrightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}^{2-}(a q)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

Short Answer

Expert verified
Equations are balanced, agents identified: (a) Cr extsubscript{2}O extsubscript{7} extsuperscript{2-} oxidizes I extsuperscript{-}; (b) MnO extsubscript{4} extsuperscript{-} oxidizes CH extsubscript{3}OH; (c) OCl extsuperscript{-} oxidizes I extsubscript{2}; (d) NO extsubscript{3} extsuperscript{-} oxidizes As extsubscript{2}O extsubscript{3}; (e) MnO extsubscript{4} extsuperscript{-} oxidizes Br extsuperscript{-}; (f) ClO extsuperscript{-} oxidizes Pb(OH) extsubscript{4} extsuperscript{2-}.

Step by step solution

01

Balance Equation (a)

Begin by writing the half-reactions for each process. For the reduction of \ \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \ \) to \ \( \mathrm{Cr}^{3+} \ \): \ \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2} \text{O} \ \] For the oxidation of \ \( \mathrm{I}^{-} \ \) to \ \( \mathrm{IO}_{3}^{-} \ \): \ \[ 6\mathrm{I}^{-} + 6\mathrm{H}_{2}\text{O} \rightarrow \mathrm{IO}_{3}^{-} + 12\mathrm{H}^{+} + 6\mathrm{e}^{-} \ \] Combine these half-reactions and balance. The balanced equation is: \ \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{I}^{-} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + \mathrm{IO}_{3}^{-} + 7\mathrm{H}_{2} \text{O} \ \] The oxidizing agent is \ \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \ \) and the reducing agent is \ \( \mathrm{I}^{-} \ \).
02

Balance Equation (b)

Write the half-reactions. For the reduction of \ \( \mathrm{MnO}_{4}^{-} \ \) to \ \( \mathrm{Mn}^{2+} \ \): \ \[ \mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\text{O} \ \] For the oxidation of \ \( \mathrm{CH}_{3}\mathrm{OH} \ \) to \ \( \mathrm{HCOOH} \ \): \ \[ \mathrm{CH}_{3}\mathrm{OH} + \mathrm{H}_{2}\text{O} \rightarrow \mathrm{HCOOH} + 4\mathrm{H}^{+} + 4\mathrm{e}^{-} \ \] Balance the electron transfer: \ \[ 4\left(\text{Reduction half-reaction}\right), 5\left(\text{Oxidation half-reaction}\right) \ \] The balanced equation is: \ \[ 4 \mathrm{MnO}_{4}^{-} + 5\mathrm{CH}_{3}\mathrm{OH} + 12\mathrm{H}^{+} \rightarrow 4 \mathrm{Mn}^{2+} + 5\mathrm{HCOOH} + 11\mathrm{H}_{2}\text{O} \ \] The oxidizing agent is \ \( \mathrm{MnO}_{4}^{-} \ \) and the reducing agent is \ \( \mathrm{CH}_{3}\mathrm{OH} \ \).
03

Balance Equation (c)

Write the half-reactions. For oxidation of \ \( \mathrm{I}_{2} \ \) to \ \( \mathrm{IO}_{3}^{-} \ \): \ \[ \mathrm{I}_{2} + 6\mathrm{H}_{2}\text{O} \rightarrow 2\mathrm{IO}_{3}^{-} + 12\mathrm{H}^{+} + 10\mathrm{e}^{-} \ \] For the reduction of \ \( \mathrm{OCl}^{-} \ \) to \ \( \mathrm{Cl}^{-} \ \): \ \[ 2\mathrm{OCl}^{-} + 4\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-} + 2\mathrm{H}_{2}\text{O} \ \] Balance the electron transfer and combine: \ \[ 5\left(\text{Reduction half-reaction}\right), 1\left(\text{Oxidation half-reaction}\right) \ \] The balanced equation is: \ \[ \mathrm{I}_{2} + 10\mathrm{OCl}^{-} + 2\mathrm{H}_{2} \text{O} \rightarrow 2\mathrm{IO}_{3}^{-} + 10\mathrm{Cl}^{-} + 12\mathrm{H}^{+} \ \] The oxidizing agent is \ \( \mathrm{OCl}^{-} \ \) and the reducing agent is \ \( \mathrm{I}_{2} \ \).
04

Balance Equation (d)

Write the half-reactions. For the reduction of \ \( \mathrm{NO}_{3}^{-} \ \) to \ \( \mathrm{N}_{2}\mathrm{O}_{3} \ \): \ \[ 2\mathrm{NO}_{3}^{-} + 6\mathrm{H}^{+} + 4\mathrm{e}^{-} \rightarrow \mathrm{N}_{2}\mathrm{O}_{3} + 3\mathrm{H}_{2}\text{O} \ \] For oxidation of \ \( \mathrm{As}_{2}\mathrm{O}_{3} \ \) to \ \( \mathrm{H}_{3}\mathrm{AsO}_{4} \ \): \ \[ \mathrm{As}_{2}\mathrm{O}_{3} + 4\mathrm{H}_{2}\text{O} \rightarrow 2\mathrm{H}_{3}\mathrm{AsO}_{4} + 4\mathrm{H}^{+} + 4\mathrm{e}^{-} \ \] Combine the balanced equations: \ \[ \mathrm{As}_{2}\mathrm{O}_{3} + 2\mathrm{NO}_{3}^{-} + 5\mathrm{H}_{2}\text{O} \rightarrow 2\mathrm{H}_{3}\mathrm{AsO}_{4} + 6\mathrm{H}^{+} + \mathrm{N}_{2}\mathrm{O}_{3} \ \] The oxidizing agent is \ \( \mathrm{NO}_{3}^{-} \ \) and the reducing agent is \ \( \mathrm{As}_{2}\mathrm{O}_{3} \ \).
05

Balance Equation (e)

In a basic solution, balance half-reactions. For the reduction of \ \( \mathrm{MnO}_{4}^{-} \ \) to \ \( \mathrm{MnO}_{2} \ \): \ \[ \mathrm{MnO}_{4}^{-} + 2\mathrm{H}_{2}\text{O} + 3\mathrm{e}^{-} \rightarrow \mathrm{MnO}_{2} + 4\mathrm{OH}^{-} \ \] For oxidation of \ \( \mathrm{Br}^{-} \ \) to \ \( \mathrm{BrO}_{3}^{-} \ \): \ \[ \mathrm{Br}^{-} + 6\mathrm{OH}^{-} \rightarrow \mathrm{BrO}_{3}^{-} + 3\mathrm{H}_{2}\text{O} + 6\mathrm{e}^{-} \ \] Balance the electrons and combine: \ \[ 2\left(\text{Reduction half-reaction}\right), 1\left(\text{Oxidation half-reaction}\right) \ \] The balanced equation is: \ \[ 2\mathrm{MnO}_{4}^{-} +\mathrm{Br}^{-} + 4\mathrm{H}_{2}\text{O} \rightarrow 2\mathrm{MnO}_{2} + \mathrm{BrO}_{3}^{-} + 4\mathrm{OH}^{-} \ \] The oxidizing agent is \ \( \mathrm{MnO}_{4}^{-} \ \), and the reducing agent is \ \( \mathrm{Br}^{-} \ \).
06

Balance Equation (f)

For a basic solution, balance half-reactions. For the reduction of \ \( \mathrm{ClO}^{-} \ \) to \ \( \mathrm{Cl}^{-} \ \): \ \[ \mathrm{ClO}^{-} + 2\mathrm{H}_{2}\text{O} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-} \ \] For the oxidation of \ \( \mathrm{Pb}( ext{OH})_{4}^{2-} \ \) to \ \( \mathrm{PbO}_{2} \ \): \ \[ \mathrm{Pb}( ext{OH})_{4}^{2-} + 2\text{OH}^{-} \rightarrow \mathrm{PbO}_{2} + 4\mathrm{e}^{-} + 2\text{H}_{2}\text{O} \ \] Balance by scaling up the reduction half-reaction, combining into the full equation. Balanced equation: \ \[ 2\mathrm{ClO}^{-} + \mathrm{Pb}( ext{OH})_{4}^{2-} \rightarrow \mathrm{PbO}_{2} + 2\mathrm{Cl}^{-} + 2\text{OH}^{-} \ \] The oxidizing agent is \ \( \mathrm{ClO}^{-} \ \) and the reducing agent is \ \( \mathrm{Pb}( ext{OH})_{4}^{2-} \ \).

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Key Concepts

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

Oxidizing Agent
In redox reactions, an oxidizing agent, also known as an oxidant, plays a crucial role. It is the substance that gains electrons from the other reactant in the reaction. Essentially, the oxidizing agent causes the oxidation of another substance while itself being reduced. This means that during the process of the reaction, the oxidizing agent undergoes a decrease in oxidation state. For example, in the given exercise,
  • the dichromate ion ( \( \text{Cr}_{2} \text{O}_{7}^{2-} \) ) is the oxidizing agent, gaining electrons to form \( \text{Cr}^{3+} \).
  • Similarly, in equation (b), \( \text{MnO}_{4}^{-} \) is another oxidizing agent, being reduced to \( \text{Mn}^{2+} \).
Understanding these agents helps in balancing and identifying the roles substances play in complex equations.
Reducing Agent
The reducing agent, or reductant, in a redox reaction is essentially the opposite of the oxidizing agent. It donates electrons to another substance, thus oxidizing itself. In simpler terms, during the reaction, the reducing agent undergoes an increase in oxidation state as it loses electrons. To view it in action, let's consider the exercises:
  • In equation (a), iodide ion ( \( \text{I}^{-} \) ) is the reducing agent, as it donates electrons to convert into \( \text{IO}_{3}^{-} \).
  • In equation (b), methanol ( \( \text{CH}_{3}\text{OH} \) ) acts as the reducing agent, being oxidized to formic acid ( \( \text{HCOOH} \) ).
Knowing the reducing agent helps in predicting the products of such reactions as well as in balancing equations.
Half-Reaction
A half-reaction is a way to display one part of a redox reaction: either the reduction or the oxidation process. These half-reactions show the number of electrons exchanged and help in balancing the overall redox equation. Let's break it down:
  • Reduction half-reaction: This reaction involves a substance gaining electrons. Think of the dichromate ion (\( \text{Cr}_{2} \text{O}_{7}^{2-} \) ) transforming into \( \text{Cr}^{3+} \) by gaining six electrons.
  • Oxidation half-reaction: Here, a substance loses electrons. An example is iodide ion (\( \text{I}^{-} \) ) changing to \( \text{IO}_{3}^{-} \) while losing electrons.
With these half-reactions, you can accurately balance redox reactions by ensuring electron loss and gain are equal.
Acidic Solution
In the world of redox reactions, an acidic solution environment is important for specific types of chemical processes. In this setting, the reaction takes place in the presence of \( \text{H}^{+} \) ions, which help to balance the equations effectively. This is crucial for balancing oxygen and hydrogen atoms when dealing with equations involving aqueous hydrogen ions (\( \text{H}^{+} \) ) and water (\( \text{H}_{2}O \) ).

Key features of an acidic solution:

  • It provides \( \text{H}^{+} \) ions that are essential for some reactions to proceed smoothly.
  • For instance, in the reduction of \( \text{MnO}_{4}^{-} \) , \(8\text{H}^{+} \) ions are involved in converting \( \text{MnO}_{4}^{-} \) into \( \text{Mn}^{2+} \) with water as a by-product.
Such conditions simplify the balancing process as \(\text{H}^{+} \) ions can account for missing hydrogen or oxygen atoms.
Basic Solution
A basic solution in redox reactions is characterized by the presence of \( \text{OH}^{-} \) ions instead of \( \text{H}^{+} \) ions. Changes to the procedure are required when dealing with reactions taking place in a basic environment. You generally introduce these \( \text{OH}^{-} \) ions to balance hydrogen and oxygen atoms efficiently.

Essential points about basic solutions:

  • In reaction (e), \( \text{OH}^{-} \) ions balance the equation for the transformation of \(\text{MnO}_{4}^{-} \) into \( \text{MnO}_{2} \).
  • This environment helps neutralize excess \( \text{H}_{2} \text{O} \) formed or required in the reactions.
Working with basic solutions necessitates reconfiguring how electrons and other ions balance within the reaction framework.

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Most popular questions from this chapter

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a magnesium strip placed in a solution of \(\mathrm{MgCl}_{2}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiCl}_{2}\). The overall cell reaction is $$ \mathrm{Mg}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{Mg}^{2+}(a q) $$ (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 magnesium electrode to the nickel electrode or from the nickel to the magnesium? (f) In which directions do the cations and anions migrate through the solution?

Consider the voltaic cell illustrated in Figure \(20.5,\) which is based on the cell reaction $$ \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) $$ Under standard conditions, what is the maximum electrical work, in joules, that the cell can accomplish if \(50.0 \mathrm{~g}\) of copper is formed?

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 h?

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable hattery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ \qquad \begin{aligned} E_{\mathrm{red}}^{\circ} &=-0.76 \mathrm{~V} \end{aligned} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Indicate whether each statement is true or false: (a) The cathode is the electrode at which oxidation takes place. (b) A galvanic cell is another name for a voltaic cell. (c) Electrons flow spontaneously from anode to cathode in a voltaic cell.
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