L07 Chem 151 Lecture Notes - Lecture 30: Primary Standard, Nernst Equation, Ferricyanide

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harlequinimpala978

13 Dec 2016

School

WUSTL

Department

University College - Chemistry

Course

L07 Chem 151

Professor

Redden

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Document Summary

(cid:3019)(cid:3041)ln (cid:4666: this can also be applied to the half-cell potentials: solutions: = 1. 10 - (cid:3019)(cid:3041)ln (cid:4666)[(cid:3027)(cid:3041)(cid:3118)+] (cid:3019)(cid:3041)ln [(cid:2870)+], red = (cid:3032)(cid:3031),(cid:3027)(cid:3041: ox = (cid:3042),(cid:3027)(cid:3041) (cid:2869)[(cid:3027)(cid:3041)(cid:3118)+]) (cid:2868) (cid:2868) (cid:3019)(cid:3041)ln (cid:4666: red = (cid:3032)(cid:3031),(cid:3048) (cid:2869)[(cid:3048)(cid:3118)+]) (cid:2868, anode (orange, oxidation): fe2+ + 2e- fe, orange = (cid:3032)(cid:2868) (cid:3019)(cid:3041)ln (cid:4666, cathode (lemon, reduction) fe2+ + 2e- fe, lemon = (cid:3032)(cid:2868) (cid:3019)(cid:3041)ln (cid:4666, = (cid:3030)(cid:3028)(cid:3047) (cid:3042)(cid:3031)(cid:3032) (cid:3028)(cid:3041)(cid:3042)(cid:3031)(cid:3032)= (cid:3019)(cid:3041)ln(cid:4672) Notice that the (cid:2868)"s will always cancel in a concentration cell! [(cid:3032)(cid:3118)+](cid:3287)(cid:3280)(cid:3288)(cid:3290)(cid:3289)(cid:4673)+(cid:3019)(cid:3041)ln (cid:4666) (cid:2869: ferricyanide [fe(cn)6]3-, ferrocyanide [fe(cn)6]4- concentration cell, concentration cells, ex: orange anode and lemon cathode with fe electrodes. = 0. 36 v: right cell: [fe(cn)6]3- + e- [fe(cn)6]4-, (cid:3034) (cid:3047) (cid:2868, left cell: [fe(cn)6]3- + e- [fe(cn)6]4-, (cid:3032)(cid:3033)(cid:3047)(cid:2868) = (cid:3030)(cid:3032)=(cid:3019)(cid:3041)ln (cid:4666)(cid:3018)(cid:3287)(cid:3280)(cid:3281) (cid:3018)(cid:3282) (cid:4667), again the value of (cid:2868) doesn"t matter in a concentration cell: if we make qright = 1, then we have: (cid:3030)(cid:3032)=(cid:3019)(cid:3041)ln((cid:3032)(cid:3033)(cid:3047))=(cid:3019)(cid:3041)ln (cid:4666)[fe(cid:4666)cn(cid:4667)6](cid:3120) (cid:3287)(cid:3280)(cid:3281) [fe(cid:4666)cn(cid:4667)6](cid:3119) (cid:3287)(cid:3280)(cid:3281)(cid:4667) curve: this curve gives us a straight line with a slope of rt/nf.

Related Questions

MC15/20 AnimationâAnalysis 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)âZn2+(aq)+2eâ

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)+2eââCu(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.

Hints

Zinc Cathode

Flow of cations

Flow of electrons

Flow of anions

Copper Cathode

Zinc Anode

Copper Anode

Reduction half-cell

Oxidation half-cell

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Correct

The cell reaction for the voltaic cell is

Zn(s)+Cu2+(aq)âZn2+(aq)+Cu(s)

Here, Zn undergoes oxidation by losing 2eâ, and the Cu2+ ions accept 2eâ 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

A standard galvanic cell is constructed with Cu²⁺ | Cu and I₂ | I^- half cell compartments connected by a salt bridge. Which of the following statements are correct?

Hint: Refer to a table of standard reduction potentials.

(Choose all that apply.)

I₂ is reduced at the cathode.

I^- is oxidized at the anode.

The anode compartment is the Cu²⁺|Cu compartment.

In the external circuit, electrons flow from the Cu²⁺|Cu compartment to the I₂|I^- compartment.

As the cell runs, anions will migrate from the Cu²⁺|Cu compartment to the I₂|I^- compartment.

L07 Chem 151 Lecture Notes - Lecture 30: Primary Standard, Nernst Equation, Ferricyanide

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