Homework solutions the half cell with the more -ve E will be the anode The way to do these is to write out both as reductions and then from info given see which is the anode --it will have the most negative E 1. I 2 + 2e 2I - Na + + e Na } E(I 2 ) > E (Na) 2. Au + + e Au I 2 + 2e 2I - } anode E(Au) > E (I) 3. Ag + + e Ag Na + + e Na } E(Ag) anode > E (Na) 22-1
Homework solutions 4. Au(s) and Ag + (aq) do not react the half cell with the more -ve E will be the anode Ag + Au + + e Ag + e Au } E (Au > E(Ag) Will not be the anode: so most +ve 5. Ag(s) and I 2 (s) do not react. Ag + + e Ag I 2 + 2e 2I - } E(Ag) > E (I) Will not be the anode: so most +ve So E(Au) > E(Ag) > E (I) > E(Na) Actually +1.69 + 0.79 +0.54-2.71 22-2
Lecture 22: Start Point WEEK 8 Lectures 22-24 Be into Chapter 20 Ebbing problems Lecture 22 Monday: Problems, Practical cells Lecture 23 Wednesday: Lecture 24 Friday Electrolysis Electrolysis 22-3
Equation Review )GE = - n F EE )G = - n F E )GE = -nfee = -RTRnK eq EE = (RT/nF).RnK eq Know all these K eq = e (nfee/rt) At 298 K 2.303RT/F = 0.0591 EE = (0.0591/n)log 10 K eq K eq = 10 (nee/0.0591) )G = )GE + RTRnQ )GE = -RTRlnK eq = -nfee )G = RTRn(Q/K) = -nfe At eq. Q/K =1 and E =0 battery dead E = EE - (RT/nF)RnQ 6 EE - (0.0591/n)log 10 Q E = -(RT/nF)Rn(Q/K) 6 -(0.0591/n)log 10 (Q/K) 22-4
review 1a. What is the voltage at 298K of the following cell? Co Co 2+ Ni 2+ Ni 1b. If [Ni 2+ ] = 0.010 M and [Co 2+ ] = 1.0 M, what is the cell voltage, the cell reaction and the cell diagram? 1a:- Get SRP values Co 2+ (aq) + 2e - 6 Co(s) -0.28 Ni 2+ (aq) + 2e - 6 Ni(s) -0.25 Standard cell is reaction is then Ni 2+ (aq) + Co(s) 6 Ni(s) + Co 2+ (aq) EE= 0.03 V remember: EE(cell) = EE(cathode) - EE(anode) SRP s 22-5
1b. Non standard cell E = EE - (0.0591/n)log 10 Q Q = [Co 2+ ]/[Ni 2+ ] = 0.03 - (0.0591/2)log 10 (1.0/0.010) = 0.03-0.06 = -0.03 V review i.e. at the specified (non-standard) concentrations the reaction Ni 2+ (aq) + Co(s) 6 Ni(s) + Co 2+ (aq) has a NEGATIVE value of E, so the spontaneous reaction (at these non-standard concentrations) is: Ni(s) + Co 2+ (aq) 6 Ni 2+ (aq) + Co(s) E = +0.03 V and the cell diagram would be: Ni Ni 2+ (0.010 M) Co 2+ Co (oxidation at the anode, reduction at the cathode). So direction of reaction can vary with concentration 22-6
At 25EC, a cell based on the reaction review Ni(s) + Cl 2 (g) (0.790 atm) 6 Ni 2+ (aq) (0.100 M) + 2CR - (aq) gives an initial voltage of 1.791 V. Calculate the value of [CR - ] in the cathode cell compartment given:- Ni 2+ (aq) + 2e - 6 Ni(s) -0.231 V Cl 2 (g) + 2e - 6 2Cl - (aq) +1.361 V Cell reaction?? Ni(s) + CR 2 (g) 6 Ni 2+ (aq) + 2CR - (aq), EE = 1.592 V E = EE - (0.0591/n)log 10 Q E = 1.791 V EE = 1.592 V n = 2 Q = [Ni 2+ ] [Cl - ] 2 / [Cl 2 ] Rearrange: E = EE - (0.0591/n)log 10 Q (0.0591/n)log 10 Q = EE - E log 10 Q = (n/0.0591)(ee - E) = (2 / 0.0591)(1.592-1.791) = -6.734 Q = 10-6.734 = 1.8 x 10-7 = (0.100). [Cl - ] 2 / 0.790 [CR - ] 2 = 1.8 x 10-7 (0.790 / 0.100) = 1.4 x 10-6 [CR - ] = 1.2 x 10-3 M 22-7
Concentration cells These beakers contain 1.0 M solutions of Cu(NO 3 ) 2, a copper electrode, and a salt bridge. If the two copper electrodes were connected together would a current flow between them? no Both of these beakers contain a copper electrode, but the concentration of Cu(NO 3 ) 2 is different. If the two copper electrodes were connected together would a current flow between them? yes a current will flow because of chemical reactions in a direction to eliminate the concentration difference. 22-8
When a current flows, what reaction occurs at this electrode? to equalize concentrations Cu 2+ is removed from solution by reduction to Cu. The Nernst equation allows us to calculate the potential of a cell as a function of concentration. Substituting values for R, T and F gives the simplified equation: Q is the concentration ratio of the ions in the cell. 22-9
cathode When the voltage is just the result of a concentration difference we can write: E = -(0.059/n) log((canode)/ccathode)) log(0.10/1.0) = log (0.10) = -1. Therefore E = -(0.059/2)(-1) = 0.030 V. This voltage is the result of the difference in Cu 2+ concentrations in the two beakers. We can use the voltage to measure the concentration of Cu 2+ ions in a solution. 22-10
Solubility products This cell has 0.10 M AgNO 3 in one beaker and 1.0 M AgNO 3 in the other. cathode E = E 0-0.0591/n log Q A piece of silver wire is used as the electrode in each beaker. What reaction occurs in the 1.0 M solution? E = 0-0.059log(0.10/1.0) = -0.059(-1) = 0.059 volts E = -(0.059/n) log((canode)/ccathode)) Another way Q =[prod]/[react] 1.0M Ag+ is used up (reacts) so Q =.1/1 =.1 22-11
What happens to the cell voltage? precipitation of AgCl decreases the concentration of Ag + which causes the voltage to increase. For example say the conc went from 1M to 0.5 M then E = -0.0591 log.1/.5 = 0.04 When a solution of NaCl is added to the AgNO3 solution, AgCl precipitates. Intuition tells us this too: as the conc in the beakers equalize the voltage would become zero. The addition of NaCl precipitated AgCl and raised the voltage to 0.40 volts. Let's calculate the concentration of Ag + in the solution containing solid AgCl. 0.40 = -(0.059/1)log(x/1.0) x= [Ag + ] = 1.6x10-7 M This means that the concentration of Ag + in equilibrium with solid AgCl is 1.6x10-7 M. 22-12
K sp (AgCl) = [Ag + ][Cl - ]= 1.6 X 10-10 From the cell voltage [Ag + ] = 1.6x10-7 M We can look up Ksp from tables and thus measure the conc of Cl - in the solution Ksp(AgCl) = [Ag + ][Cl - ]=[1.6x10-7 ] [Cl - ] = 1.6x10-10 [Cl - ] = 1.0x10-3 M 22-13
Determination of K sp for an insoluble salt. Given: 1. AgBr(s) + e - 6 Ag(s) + Br - (aq) EE = +0.071 2. Ag + (aq) + e - 6 Ag(s) EE = +0.799 V Take equation 1, reverse 2 and add: AgBr(s) + e - 6 Ag(s) + Br - (aq), EE = +0.071 Ag(s) 6 Ag + (aq) + e - EE = -0.799 V AgBr(s) º Ag + (aq) + Br - (aq), EE = -0.728 V note the negative value. What does this tell you? Remember that if you know any one of )GE, K eq or EE, you know all three! K eq = 10 (nee/0.0591) K sp = 10 {(1 x -0.728)/0.0591} = 10-12.3 = 5 x 10-13 M 2 22-14
ph electrodes and ph meters A ph meter is simply an ion-selective electrode with a membrane which is a very thin piece of special glass which is permeable to H + ion. Inside the membrane is a solution of a HCR of fixed concentration. A reference electrode completes the circuit. The voltage of such a system is given by the equation: E = constant - 0.0591 x log {1/[H + ]} = constant + 0.0591 x ph What this means is that a ph meter will change its potential by 0.0591 V (59.1 mv) for each change of 1 ph unit. The read-out voltmeter is set to read directly in ph units. The meter is calibrated with a buffer of known ph before use. [Note the logarithmic scale.] 22-15
Homework Predict the effect on the cell voltage (Bigger, Smaller, No Change) from a cell with the reaction: 2H + (aq) + SO 4 2- + Pb(s) º H 2 (g) + PbSO 4 (s) (a) Increase in pressure of H 2 (g) (b) Increase in size of the Pb electrode (c) Decrease in ph (d) Dilution of the electrolyte with water (e) Addition of Na 2 SO 4 to the cell electrolyte (f) Decrease in amount of PbSO 4 (s) (g) Addition of a small amount of NaOH E = 0.360 - (0.0591/2) x log{ph 2 (g)/[h + ] 2 [SO 4 2- ]} Anything that makes the reaction go L 6 R gives a bigger cell voltage. 22-16