CHEM 001C Lecture Notes - Lecture 15: Standard Electrode Potential, Electromotive Force, Membrane Potential

1. Consider a voltaic (galvanic) cell with the following metal electrodes. Identify which metal is the cathode and which is the anode, and calculate the cell potential.

(a) Al and Co(II)
Cathode: ____

Anode: ______

Ecell =

(b) Cd(II) and Ag(I)

cathode: ___

anode: ___

Ecell= ____

(c) Cr(III) and Sc(III)

cathode: ___

Anode:_____

Ecell:____

2. A voltaic cell contains two half-cells. One half-cell contains a titanium electrode immersed in a 1.00 M Ti(NO₃)₃ solution. The second half-cell contains a zinc electrode immersed in a 1.00 M Zn(NO₃)₂ solution.

(a) Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell.

(b) Write the overall balanced equation for the voltaic cell. (Include states-of-matter under the given conditions in your answer.)

3. ΔG° and E° can be said to measure the same thing, and are convertible by the equation

ΔG° = −nFE⁰cell 

where n is the total number of moles of electrons being transferred, and F is the Faraday constant 9.64853415 ✕ 10⁴ C/mol. The free energy (ΔG°) of a spontaneous reaction is always negative.

For each of the electrochemical cells below, calculate the free energy of the system and state whether the reaction is spontaneous or non-spontaneous as written based on the cathode and anode assignment given. (Use the table of Standard Electrode Potentials.)

(a) The cathode is Zn(II) and the anode is Co(II).

free energy: ____ kJ

spontaneity: _____

MC15/20 Animation—Analysis of Copper-Zinc Voltaic Cell

Some oxidation-reduction reactions are spontaneous, and the energy released by them can be used for electrical work. This principle is used in the working of a voltaic cell. Voltaic cells, or electrochemical cells, make use of the electrons that are transferred from the species that releases electrons and undergoes oxidation to the species that accepts electrons and undergoes reduction. If it is possible by any means to separate the oxidation reaction and the reduction reaction and then connect them externally so that the electrons flow from one compartment to the other, you can construct an electrochemical cell.

Watch the video that describes the cell reaction and the cell components of a voltaic cell.

In a copper-zinc voltaic cell, one half-cell consists of a Zn electrode inserted in a solution of zinc sulfate and the other half-cell consists of a Cu electrode inserted in a copper sulfate solution. These two half-cells are separated by a salt bridge.

At the zinc electrode (anode), Zn metal undergoes oxidation by losing two electrons and enters the solution as Zn2+ ions. The oxidation half-cell reaction that takes place at the anode is

Zn(s)→Zn2+(aq)+2e−

The Cu ions undergo reduction by accepting two electrons from the copper electrode (cathode) and depositing on the electrode as Cu(s). The reduction half-cell reaction that takes place at the cathode is

Cu2+(aq)+2e−→Cu(s)

The electrons lost by the Zn metal are gained by the Cu ion. The transfer of electrons between Zn metal and Cu ions is made possible by connecting the wire between the Zn electrode and the Cu electrode. Thus, in the voltaic cell, the electrons flow through an external circuit from the anode to the cathode. For a voltaic cell to work, the solution in the two half-cells must remain electrically neutral. This can happen only if the flow of ions is countered with the flow of electrons. The flow of ions is made possible with the use of a salt bridge. A salt bridge is a solution of some other metal that has common ions. If a copper-zinc voltaic cell utilizes ZnSO4 and CuSO4solution, you will use a saturated Na2SO4 solution in the salt bridge. Thus, the salt bridge will help the migration of ions across the two compartments, or two half-cells.

Part A

Label the diagram according to the components and processes of a voltaic cell.

Drag the appropriate labels to their respective targets.

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Correct

The cell reaction for the voltaic cell is

Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)

Here, Zn undergoes oxidation by losing 2e−, and the Cu2+ ions accept 2e− to form metallic Cu, which is deposited at the copper electrode. At the zinc electrode, oxidation occurs, and hence it is known as the anode. Reduction occurs at the copper electrode, and hence it is called the cathode.

The electrochemical cell notation for this cell is Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s).

Construction of a voltaic cell

Consider a chromium-silver voltaic cell that is constructed such that one half-cell consists of the chromium, Cr, electrode immersed in a Cr(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Cr electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to solve Parts B, C, and D.

Part B

The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell is the site of the reduction reaction.

Type the half-cell reaction that takes place at the anode for the chromium-silver voltaic cell. Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.

Express your answer as a chemical equation.

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Part C

The half-cell is a chamber in the voltaic cell where one half-cell is the site of an oxidation reaction and the other half-cell is the site of a reduction reaction.

Type the half-cell reaction that takes place at the cathode for the chromium-silver voltaic cell. Indicate physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.

Express your answer as a chemical equation.

Hints

You are asked to construct a voltaic cell that would simulate an alkaline battery by providing an electrical output of 1.50 VV at the beginning of its discharge. After you complete it, your voltaic cell will be used to power an external device that draws a constant current of 0.50 AA (amperes) for 2.0 hh. You are given the following supplies: electrodes of each transition metal from manganese to zinc; the chloride salts of the +2 transition metal ions from Mn2+Mn2+ to Zn2+Zn2+ (MnCl2MnCl2, FeCl2FeCl2, CoCl2CoCl2, NiCl2NiCl2, CuCl2CuCl2, and ZnCl2ZnCl2); two 100 mLmL beakers; a salt bridge; a voltmeter; and wires to make electrical connections between the electrodes and the voltmeter. The standard reduction potentials for the divalent metal ions are as follows: Half-reactionMn2+(aq)Fe2+(aq)Co2+(aq)Ni2+(aq)Cu2+(aq)Zn2+(aq)++++++2e−2e−2e−2e−2e−2e−→→→→→→Mn(s)Fe(s)Co(s)Ni(s)Cu(s)Zn(s)E∘(V)−1.18−0.44−0.28−0.250.34−0.76 Since you were tasked with constructing a cell that produces a potential of 1.50 VV, you decide to fine tune your concentrations to minimize the error. If you used the same concentration for each electrolyte, your voltaic cell would produce a potential of 1.52 VV. Therefore, you should adjust the concentration ratio that would lower your cell potential by 0.02 VV. You and a fellow researcher decide that you should prepare two 100 mLmL solutions for the voltaic cell. Your fellow researcher already prepared one of the electrolyte solutions for you: 0.228 MM CuCl2 CuCl2. At what concentration should you prepare the MnCl2MnCl2 solution such that your initial voltaic cell potential is 1.50 VV? Assume that the temperature is consistently 25 ∘C∘C. If the battery is permitted to discharge for a period of two hours at a constant current of 0.50 AA, what will be the concentration of the manganese(II) ion in the anode electrolyte solution at the end of the discharge? Similarly, what will be the concentration of the copper(II) ion in the cathode electrolyte solution at the end of the discharge? Given the nature of how a cell is constructed, you would usually deplete the ions at the cathode before completely oxidizing the anode (since the latter would break the circuit). This is because the complete oxidation of the anode would likely only occur if you were using an extremely thin wire of the metal. In this case, let us assume you constructed the voltaic cell using sufficiently sized rods for each metal. This means that the total time you can run your cell will depend on the initial copper ion concentration. Calculate the maximum run time for your voltaic cell if it runs at 0.5 AA and your initial concentration of CuCl2(aq)CuCl2(aq) was 0.228 M'