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Chapter 17: Problem 22

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

Short Answer

Expert verified
A galvanic cell can be constructed from two substances, each having a negative standard reduction potential, by selecting the half-reaction with the higher potential as the cathode and the other as the anode. Ensure that the overall cell potential (E掳cell) is positive for the reaction to be spontaneous. Construct the cell by placing the substances in separate beakers with respective metal strips and connecting them with a wire and a salt bridge or porous barrier. The redox reaction will spontaneously generate an electric current once connected.

Step by step solution

01

Understand the galvanic cell and reduction potentials.

A galvanic cell, also commonly known as a voltaic cell, is an electrochemical cell that uses a spontaneous redox (reduction-oxidation) reaction to generate an electric current. A redox reaction consists of two half-reactions, where one substance undergoes reduction, and another undergoes oxidation. Reduction potentials are values used to predict which half-reaction will proceed as a reduction reaction (gain of electrons) and which will proceed as an oxidation reaction (loss of electrons) when substances are combined in a galvanic cell. A higher standard reduction potential substance will have a greater tendency to gain electrons. When comparing the reduction potentials of two substances (E掳 values), the reaction becomes spontaneous if the final E掳 is positive. In this exercise, both substances have negative standard reduction potentials. Let's call these two substances A and B and their respective standard reduction potentials E掳(A) and E掳(B).
02

Determine the half-reaction with the higher potential.

Compare the standard reduction potential of substance A and substance B. Select the one with the higher reduction potential. Let's assume that E掳(A) > E掳(B), so substance A will be the cathode where the reduction reaction will occur, and substance B will be the anode where the oxidation reaction will occur. Cathode half-reaction: A + ne^鈭� 鈫� A(reduced form) Anode half-reaction: B 鈫� B(oxidized form) + ne^鈭�
03

Calculate the overall cell potential and determine spontaneity.

The cell potential for the spontaneous reaction (E掳cell) is given by the difference in the standard reduction potentials of the two half-reactions: E掳cell = E掳(cathode) - E掳(anode) Evaluate E掳cell by substituting the E掳 values of the two substances A and B: E掳cell = E掳(A) - E掳(B) Since both E掳(A) and E掳(B) are negative, and E掳(A) is greater than E掳(B), the E掳cell will be positive, and the reaction will be spontaneous.
04

Construct the galvanic cell.

Construct a galvanic cell using the following steps: 1. Place substance A in a beaker and immerse a clean metal strip (such as platinum) into it as the cathode. 2. Place substance B in a separate beaker and immerse a clean metal strip (such as platinum) into it as the anode. 3. Connect the metal strips with a wire, and a salt bridge or a porous barrier should be used to connect the two solutions allowing the movement of ions while blocking the direct flow of solutions. 4. When the cell is connected, the redox reaction will spontaneously start to generate an electric current through the wire. In this way, one can construct a galvanic cell from two substances, each having a negative standard reduction potential.

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Most popular questions from this chapter

Which of the following statements concerning corrosion is(are) true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) \(\mathbf{c} . \operatorname{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q)\) d. \(\mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Bawe }}{\longrightarrow}\) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Barc}}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)Balance the following equations by the half- reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) \(\mathbf{c} . \operatorname{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q)\) d. \(\mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow}\) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Base}}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)

What reactions take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{NiBr}_{2}\) b. molten \(\mathrm{AlF}_{3}\) c. molten \(\mathrm{MnI}_{2}\)
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