Electrochemical cells

Transcription

1 Electrchemical cells In this chapter, we turn ur attentin t electrn transfer reactins. T identify an electrn transfer reactins, we must assign xidatin states review rules in Chapter 3. e.g. Zn(s) Cu(NO 3 ) 2 (aq) A cpper cating frms spntaneusly n the zinc rd because f the fllwing reactin. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) All xidatin-reductin reactins can be split int tw separate halfreactins. xidatin: reductin: verall: In this particular example, the electrn transfer is direct in the sense that the reactants are in direct cntact with each ther. The interesting (and useful) thing abut electrn transfer reactins is that they can be carried ut even if the reactants are nt in direct cntact!

2 Electrchemical cells the reactants are physically separated but are electrically cnnected by a cnducting material made up f tw half-cells and a salt bridge The reactin between Zn(s) and Cu 2+ (aq) culd be carried ut using the fllwing set up. Electrns flw thrugh cnducting materials (e.g. Pt wire, Cu wire, etc.) An electrde immersed in a slutin. An electrde is a slid that prvides the surface fr the electrn transfer. A tube filled with cncentrated salt slutin. platinum wire Zn(s) salt bridge Cu(s) Zn(NO 3 ) 2 (aq) Cu(NO 3 ) 2 (aq) In this example, the zinc electrde is the ande and the cpper electrde is the cathde. ande = electrde where xidatin ccurs (left) cathde = electrde where reductin ccurs (right) An electrchemical cell can be classified as: (1) galvanic r vltaic based n a spntaneus redx reactin Galvanic cell prduces energy a battery is a galvanic cell (2) electrlytic based n a nn-spntaneus redx reactin Electrlytic cell requires/cnsumes energy electrlytic cells are imprtant fr cnverting metal ins int metallic slids)

3 Chemists have devised a shrthand ntatin fr describing the cmpnents making up an electrchemical cell. Using this ntatin, the electrchemical cell discussed previusly is represented as: Zn(s) Zn(NO 3 ) 2 (aq) Cu(NO 3 ) 2 (aq) Cu(s) What is the functin f the salt bridge? The salt bridge prevents charge build-up in the electrchemical cell. Cnsider what happens in ur electrchemical cell if the salt bridge is nt there. Zn(s) platinum wire Cu(s) We wuld get a build-up f psitive charge in ne slutin and a build-up f negative charge in the ther slutin. Zn(NO 3 ) 2 (aq) Cu(NO 3 ) 2 (aq) With the salt bridge in place, the psitive ins in the salt bridge migrate twards ne end f the tube and sme f them diffuse int the slutin t ffset the build up f negative charge that wuld therwise ccur in that slutin. The negative ins in the salt bridge travel in the ppsite directin. By the end f this unit, yu shuld be able t predict fr any cell the directin f electrn flw and the directins f in migratin in the salt bridge.

4 The Cell Ptential The cell ptential is an imprtant quantity t cnsider when dealing with electrchemical cells. cell ptential = The cell ptential depends n: (i) the xidizing r reducing strength f each half-cell (ii) the cnditins under which the cell perates Our first cncern is t investigate the xidizing/reducing strengths f different half cells. In rder t cme up with a scale that ranks the xidizing/reducing strengths f half-cells quantitatively, we must fix the cnditins Standard cnditins cmpare all pssible half-cells t an accepted reference half-cell

5 Standard hydrgen electrde (SHE) and standard electrde ptentials The basic cnstructin f the standard hydrgen electrde is shwn belw. H 2 (g) 1 atm H + (aq) 1 ml L 1 The half-reactins assciated with the SHE are: 2 H + (aq) + 2 e H 2 (g) If the SHE is the CATHODE. H 2 (g) 2 H + (aq) + 2 e If the SHE is the ANODE. These tw half-reactins are assigned a vltage f zer vlts (fr standard cnditins). We can use the fllwing strategy t assign a vltage t ther half-reactins. (1) Cnnect anther half-cell t the SHE (2) Measure the cell ptential (i.e. vltage) and determine the directin f electrn flw. (3) Assign that vltage t the half-cell reactin using a + sign r a sign.

6 Table f Standard Reductin Ptentials Half-reactin E / V F 2 (g) + 2e 2 F (aq) Au + (aq) + e Au(s) MnO 4 (aq) + 8H + (aq) + 5e Mn 2+ (aq) + 4H 2 O(l) O 2 (g) + 4H + (aq) + 4e 2H 2 O(l) Br 2 (l) + 2e 2Br (aq) Ag + (aq) + e Ag(s) Fe 3+ (aq) + e Fe 2+ (aq) 0.77 O 2 (g) + 2H 2 O(l) + 4e 4OH (aq) Cu 2+ (aq) + 2e Cu(s) AgCl(s) + e Ag(s) + Cl (aq) 0.22 AgBr(s) + e Ag(s) + Br (aq) H + (aq) + 2e H 2 (g) Fe 3+ (aq) + 3e Fe(s) 0.04 Pb 2+ (aq) + 2e Pb(s) Ni 2+ (aq) + 2e Ni(s) C 2+ (aq) + 2e C(s) Tl + (aq) + e Tl(s) Cd 2+ (aq) + 2e Cd(s) Fe 2+ (aq) + 2e Fe(s) 0.41 Cr 3+ (aq) + 3e Cr(s) 0.74 Zn 2+ (aq) + 2e Zn(s) Na + (aq) + e Na(s) Li + (aq) + e Li(s) 3.00 Nte carefully: all f the half-reactins in the table are written as reductins the E value assigned t each half-reactin is the vltage we d get if a halfcell based n that half-reactin was cnnected t the SHE the mre psitive the E value, the greater the tendency f the reactin t ccur as written when cupling tw half-reactins, the half-reactin with the mre psitive E value has a greater tendency t ccur as a reductin.

7 Using standard reductin ptentials t calculate E cell The cell ptential is equal t the energy transferred per unit f charge transferred: With this in mind, it is straightfrward t establish the fllwing rules. If we reverse a half-reactin, then E changes sign. If we multiply a half-reactin by n, then E is unchanged. If we add tw half-reactins tgether, then we add the tw E values. We can use these rules, and a table f standard reductin ptentials, t calculate E cell fr any electrchemical cell.

8 Example: A galvanic cell is cnstructed using Pb/Pb 2+ and Ag/Ag + halfcells. The cell perates under standard cnditins. (a) Identify the ande and the cathde. (b) Draw a diagram f the cell. What is the shrthand ntatin fr the cell? Shw the directin f electrn transfer. Which electrde increases in mass? (c) What is? E cell

9 A different way t calculate E cell? Nt really In the previus example, we calculated E cell using the equatin belw. E = E + E cell xid red When we use the equatin abve, we are emphasizing that the verall reactin is the sum f an xidatin prcess and a reductin prcess, and that the cell ptential is the sum f an xidatin ptential and a reductin ptential. Yur text bk uses the fllwing equatin t calculate the cell ptential: In the equatin abve, E = E E cell cathde ande. E cathde = The reductin ptential fr the cathde reactin (written as a reductin) E ande = The reductin ptential fr the ande reactin (written as a reductin) When we use the equatin E = E E cell cathde ande, we are emphasizing that E cell depends n the difference between the reducing strengths f the tw half-cells and that E cell is the difference between tw ptentials. reductin Either ne f these equatins can be used t calculate E cell, prvided yu use the equatins prperly and understand the meaning f the terms invlved.

10 What des the sign f E tell us? Armed with the frmulae E = E + E cell xid red cell cathde ande and E = E E we can calculate the value f E fr any xidatin-reductin reactin prvided we knw the reductin ptentials fr the tw half-reactins. The E value fr a given reactin can be psitive r negative, and the sign f the E value has the fllwing interpretatin. (We will justify this interpretatin later.) E > 0 Under standard cnditins, the reactin prceeds spntaneusly in the frward directin E < 0 Under standard cnditins, the reactin prceeds spntaneusly in the reverse directin E = 0 Under standard cnditins, the reactin is at equilibrium.

11 Oxidizing and reducing agents A table f standard reductin ptentials can be used t identify xidizing and reducing agents. xidizing agent causes smething else t be xidized a gd xidizing agent is a substance that is easily reduced reducing agent causes smething else t be reduced a gd reducing agent is a substance that is easily xidized Frm the table f standard reductin ptentials, we can see that: F 2 (g) is a strng xidizing agent because F 2 (g) has a strng tendency t be reduced i.e. it has a very psitive reductin ptential Na(s) is a strng reducing agent because Na(s) has a very psitive xidatin ptential r a very negative reductin ptential

12 Example: Which f the fllwing substances mst readily reduces Pb 2+ t Pb under standard cnditins? Ag + (aq) H 2 (g) Fe 3+ (aq) Br (aq) Zn(s) Li + (aq)

13 The Nernst Equatin The Nernst equatin can be used t understand hw the cell ptential changes with cncentratin. E = V E lg Q n The Nernst equatin is btained when yu apply the laws f thermdynamics t electrchemical cells. The derivatin f this equatin requires discussin f the three laws f thermdynamics. Unfrtunately, a meaningful discussin f the laws f thermdynamics is a curse in itself, e.g. CHEM 254. We dn t expect yu t knw where this equatin cmes frm, but we d expect yu t knw hw t use it. The Nernst equatin will be supplied n the Data Sheet. Imprtant pints t remember (1) Q in the equatin abve is the thermdynamic reactin qutient. The Thermdynamic Reactin Qutient Fr a gas-phase reactin, the thermdynamic reactin qutient Q is identical t Q p. N 2 (g) + 3 H 2 (g) º 2 NH 3 (g) (2) Under standard cnditins, Q = 1 and s E = E V lg 1 n Q = Q p = 2 PNH 3 3 P N2 H2 Fr a reactin that ccurs in aqueus slutin, Q is identical t Q c. P Zn(s) + Cu 2+ (aq) º Zn 2+ (aq) + Cu(s) (3) The Nernst equatin can be applied t the verall reactin r t halfreactins individually. [ 2+ Zn ] Q = Q c = [ ] 2+ Cu Q is different frm Q p r Q c when the reactin invlves gases and aqueus slutins: H 2 (g) + Cu 2+ (aq) º 2 H + (aq) + Cu(s) Q = + [ H ] 2+ [ Cu ] 2 PH 2

14 Example: What is E fr the half-reactin 2 H + (aq) + 2e º H 2 (g) if [H + ] = 0.10 ml L 1 and P H2 = 1.5 atm? Example: Calculate E cell fr the fllwing galvanic cell. Cu(s) Cu 2+ (aq, 1.5 ml L 1 ) Fe 3+ (aq, 1.5 ml L 1 ), Fe 2+ (aq, 0.50 ml L 1 ) Pt(s)

15 Electrchemical cells and equilibrium In the previus example, we cnsidered the fllwing electrchemical cell. platinum wire Cu(s) salt bridge [Fe 3+ ] = 0.5 ml L 1 [Fe 2+ ] = 1.5 ml L 1 [Cu 2+ ] = 1.5 ml L 1 We calculated that E cell = V and predicted that electrns wuld flw spntaneusly frm the cpper electrde twards the Fe 2+ /Fe 3+ slutin. Cu(s) is spntaneusly xidized t Cu 2+ Fe 3+ is spntaneusly reduced t Fe 2+

16 Hw des the reactin in the cell affect the values f Q and E cell? Shuld we expect E cell t decrease indefinitely? N! We knw that a clsed system prceeds spntaneusly twards a state f dynamic equilibrium. When equilibrium is reached, there is n further cnversin f reactants int prducts. When an electrn transfer reactin stps, there is n current flwing and the vltage must be zer! At equilibrium:

17 Example: Use standard reductin ptentials t calculate the equilibrium cnstant fr the reactin belw. 2 Cr 3+ (aq) + 3 Ni(s) º 2 Cr(s) + 3 Ni 2+ (aq)

18 Example: Use standard reductin ptentials t calculate the equilibrium cnstant fr the reactin belw. AgBr(s) º Ag + (aq) + Br (aq)

19 Electrlytic cells and electrlysis Electrlytic cells use an external vltage surce t frce a nnspntaneus redx reactin t ccur. Electrlytic cells are useful fr cnverting xidized metals back int metallic frm. e.g. Na + (aq) + e Na(s) E = 2.71 V In rder t reduce prduce Na(s) frm Na + (aq), we wuld have t use an external pwer surce capable f supplying at least 2.71 vlts.

20 Example: A cpper sulfate slutin, CuSO 4 (aq), is electrlyzed fr 7.00 minutes using an external current f 0.60 amperes. What mass f Cu(s) is prduced? (Nte: 1 ampere = 1 culmb per secnd) Recall: 1 ml e C The cathde reactin is: ande Cu(s) cathde Cu 2+ (aq) + 2e Cu(s) E red = V The battery draws electrns away frm the ande. If the ande is chsen t be an inert material (e.g. Pt), then what reactin is supplying these electrns? The xidatin f water!! Cu(NO 3 ) 2 (aq) 2 H 2 O(l) O 2 (g) + 4 H + (aq) + 4 e E xid = 1.23 V Withut the battery, E cell = V. In rder fr Cu2+ t be reduced, we need a battery with a vltage significantly greater than V.