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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°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° values), the reaction becomes spontaneous if the final E° 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°(A) and E°(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°(A) > E°(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^− → A(reduced form) Anode half-reaction: B → B(oxidized form) + ne^−
03

Calculate the overall cell potential and determine spontaneity.

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

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

The overall reaction in the lead storage battery is $$\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ a. For the cell reaction \(\Delta H^{\circ}=-315.9 \mathrm{kJ}\) and \(\Delta S^{\circ}=\) \(263.5 \mathrm{J} / \mathrm{K} .\) Calculate \(\mathscr{E}^{\circ}\) at \(-20 .^{\circ} \mathrm{C} .\) Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. b. Calculate \(\mathscr{E}\) at \(-20 .^{\circ} \mathrm{C}\) when \(\left[\mathrm{HSO}_{4}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.5 \mathrm{M}\) c. Consider your answer to Exercise \(71 .\) Why does it seem that batteries fail more often on cold days than on warm days?

Consider the standard galvanic cell based on the following half-reactions: $$\begin{array}{c} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\mathrm{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\left.\mathrm{Cu}^{2+} \text { reacts with } \mathrm{NH}_{3} \text { to form } \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\left.\mathrm{Ag}^{+} \text {reacts with } \mathrm{Cl}^{-} \text {to form } \mathrm{AgCl}(s) .\right]\) d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$ \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V} $$

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. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

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.

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