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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
To construct a galvanic cell using two substances with negative standard reduction potentials, first compare the standard reduction potentials of the two substances. The substance with a lower (more negative) standard reduction potential will be oxidized and act as the anode, while the substance with a higher (less negative) standard reduction potential will be reduced and act as the cathode. Construct two half-cells containing the anode substance and its electrolyte, and the cathode substance and its electrolyte. Connect the half-cells with a salt bridge allowing ion flow and connect the electrodes with an external wire allowing electron flow. The constructed galvanic cell can then generate electrical energy through the redox reaction between the two substances.

Step by step solution

01

Understand the concept of galvanic cells

Galvanic cells, also known as voltaic cells or electrochemical cells, are devices that use spontaneous redox (reduction-oxidation) reactions to generate electrical energy. They have two half-cells consisting of an electrode and an electrolyte. One half-cell involves the oxidation process and the other involves the reduction process. The flow of electrons from the anode (where oxidation occurs) to the cathode (where reduction occurs) generates the electrical energy.
02

Learn about standard reduction potentials

Standard reduction potential is a measure of how likely a substance is to be reduced in an electrochemical reaction. It is measured in volts (V) and is defined under standard conditions (1 M concentration of each ion, 1 atm pressure, 25°C). A substance with a positive standard reduction potential is more likely to be reduced, while a substance with a negative standard reduction potential is less likely to be reduced. The substance with the higher (i.e., less negative or more positive) standard reduction potential will tend to be reduced (gain electrons), while the substance with the lower standard reduction potential will tend to be oxidized (lose electrons).
03

Choose the two substances with negative standard reduction potentials

For this exercise, let's consider two substances, X and Y, with negative standard reduction potentials. Let the standard reduction potential for X (E°₁) be -0.30 V, and for Y (E°₂), -0.60 V. These values indicate that both substances have a negative reduction potential, and we need to determine the direction of the reaction in the constructed galvanic cell.
04

Determine the anode and cathode in the galvanic cell

To determine which substance becomes the anode (where oxidation occurs) and which becomes the cathode (where reduction occurs) in the galvanic cell, compare their standard reduction potentials. The substance with the lower (more negative) standard reduction potential will be oxidized (lose electrons) and act as the anode. The substance with the higher (less negative) standard reduction potential will be reduced (gain electrons) and act as the cathode. In this case, substance Y has a lower standard reduction potential (E°₂ = -0.60 V) than substance X (E°₁ = -0.30 V). Therefore, Y will be the anode and X will be the cathode.
05

Construct the galvanic cell

To construct the final galvanic cell: 1. Build two half-cells. In one half-cell, place the anode substance Y and its electrolyte - Y's ions mixed in water. In the other half-cell, place the cathode substance X and its electrolyte - X's ions mixed in water. 2. Connect the two half-cells using a salt bridge - a U-shaped tube filled with an ionic solution such as KNO₃ or KCl - which allows ions to pass between the half-cells, maintaining charge neutrality. The salt bridge balances the charges in the half-cells and completes the electrical circuit. 3. Connect the two electrodes (anode and cathode) using an external wire, allowing the flow of electrons from the anode (Y) to the cathode (X). The galvanic cell is now constructed and can generate electrical energy through the redox reaction between substance X and substance Y.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions at \(25^{\circ} \mathrm{C}\) : $$\begin{aligned}\mathrm{Ag}^{+}+\mathrm{e}^{-} & \longrightarrow \mathrm{Ag} \\\ \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ Predict whether \(\mathscr{G}_{\text {cell }}\) is larger or smaller than \(\mathscr{C}_{\text {cell }}^{\circ}\) for the following cases. a. \(\left[\mathrm{Ag}^{+}\right]=1.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) b. \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\)

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$\mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co}$$ The overall reaction and equilibrium constant value are $$2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow{2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s)} \quad K=2.79 \times 10^{7}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

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.

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

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