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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 A and B with negative standard reduction potentials, first determine the anode (Substance A) and cathode (Substance B) based on their reduction potentials. Then, set up the cell with the notation: Substance A | A^n+ || B^n+ | Substance B. Fill two containers with electrolyte solutions of A and B, place electrodes made of each substance in their respective containers, and connect them with a conducting wire including a voltmeter. Finally, connect the containers using a salt bridge to maintain charge balance.

Step by step solution

01

Identify the substances and their standard reduction potentials

For the sake of this example, let's assume we are given two substances, A and B, with their standard reduction potentials: Substance A has a standard reduction potential Eº_A of -1.2 V, and Substance B has a standard reduction potential Eº_B of -0.8 V.
02

Determine the anode and cathode

To form a spontaneous redox reaction, the substance with the higher reduction potential should be the cathode, and the substance with a lower reduction potential should be the anode. In this case, Eº_B (-0.8 V) is higher than Eº_A (-1.2 V). Therefore, Substance B will act as the cathode, and Substance A will be the anode for this galvanic cell.
03

Describe the cell notation

Now that we've identified the anode and cathode, we can write the complete cell notation. The cell notation for this galvanic cell is given by: Anode | Anode electrolyte || Cathode electrolyte | Cathode For our example: Substance A | A^n+ || B^n+ | Substance B
04

Set up the galvanic cell

To set up the galvanic cell, follow these steps: 1. Place two containers (beakers) side-by-side, one for each half-cell. 2. Fill each container with an electrolyte solution that contains the ions of Substance A and Substance B, respectively. 3. Place an electrode made of Substance A in the container with its electrolyte and an electrode made of Substance B in the container with its electrolyte. 4. Connect the two electrodes with a conducting wire, including a voltmeter in the circuit to measure the cell potential. 5. Connect the two containers via a salt bridge (a U-shaped tube filled with an inert electrolyte). This allows the flow of ions between the half-cells and helps in maintaining charge neutrality of the solutions. Now you have successfully constructed the galvanic cell using two substances with negative standard reduction potentials.

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Most popular questions from this chapter

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$\mathscr{E}_{\text { meas }}=\mathscr{E}_{\text { ref }}+0.05916 \mathrm{pH}$$ where \(\mathscr{E}_{\text { ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that \(\mathscr{E}_{\text { meas }}=0.480 \mathrm{V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the nncertainty in the measured potential is \(+1 \mathrm{mV}\) \(( \pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{KF} \quad\) b. molten \(\mathrm{CuCl}_{2} \quad\) c. molten \(\mathrm{MgI}_{2}\)

It takes 15 kWh (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is 10.7 \(\mathrm{kJ} / \mathrm{mol}(1 \text { watt }=1 \mathrm{J} / \mathrm{s}) . ]\)

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

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