Electrochemistry

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Lab-Calculations Report Sheet

A1=-0.518, A2 = + 0.524, B1= -0.862, B2= +0.538, C1= + 0.398, D1 = +1.585

Half-cell equation

E(V)

Mg(s) Mg2+(aq) + 2e

Ag+(aq) + e Ag(s)

+2.08V

Pb(s) Pb2+(aq) + 2e

Zn2+(aq) + 2e Zn(s)

+0.412V

Pb(s) Pb2+(aq) + 2e

Fe2+(aq) + 2e Fe(s)

+0.156V

Sn(s) Sn2+(aq) + 2e

Fe2+(aq) + 2e Fe(s)

+ 0.030 V

Cu(s) Cu2+ (aq)+ 2e

Ag+(aq) + eAg(s)

-0.518V

Cu(s) Cu2+(aq) + 2e

Zn2+(aq) + 2e Zn(s)

-0.862V

TheLiterature values

Half-cell equation

E(V)

Mg(s) Mg2+(aq) + 2e

Ag+(aq) + e Ag(s)

+1.58V

Pb(s) Pb2+(aq) + 2e

Zn2+(aq) + 2e Zn(s)

+0.63V

Sn(s) Sn2+(aq) + 2e

Fe2+(aq) + 2e Fe(s)

+ 0.55 V

Pb(s) Pb2+(aq) + 2e

Fe2+(aq) + 2e Fe(s)

+0.17V

Cu(s) Cu2+ (aq)+ 2e

Ag+(aq) + eAg(s)

-0.38V

Cu(s) Cu2+(aq) + 2e

Zn2+(aq) + 2e Zn(s)

-0.6V

Calculationof the Potentials

MgMg2+//Ag+Ag,=2.28+-.80 = +.58V

PbPb2+//Zn2+Zn= -0.13+.76= +0.63

PbPb2+//Sn2+Sn=-0.40+0.15= 0.55V

CuCu2+//Ag+Ag=0.16+0.80= +0.17V

CuCu2+//Zn2+Zn=0.16-0.76=-0.38V

MgMg2+//Ag+Ag= 0.16+0.8=-0.6V

Part1, section 2

Line notation for electrochemical cell

Ecell (EXP) V

Ecell (Theor) V

MgMg2+//Ag+ Ag

+2.08V

+1.58V

PbPb2+//Zn2+Zn

+0.63V

+0.63V

PbPb2+//Sn2+Sn

+ 0.030 V

+ 0.55 V

CuCu2+//Ag+ Ag

+0.156V

+0.17V

CuCu2+//Zn2+Zn

-0.38V

-0.38V

MgMg2+//Ag+ Ag

-0.862V

-0.6V

ECell=EExp– (0.0592/n) *log Q

  1. Ecell= +2.08-(0.0592/1)*log (0.1/1)

Ecell=2.1392V

  1. Ecell= 0.63-(0.0592/1)*log 0.1= 0.6892V

  2. Ecell= 0.55-(0.0592/1)*log 0.1= 0.6092V

  3. Ecell= 0.17-(0.0592/1)*log 0.1= 0.2292V

  4. Ecell= -0.38-(0.0592/1)*log 0.1= -0.3208V

  5. Ecell= -0.862V-(0.0592/1)*0.1= -0.8028V

Thecalculated values were different from the literature value because ofexperimental error that could have raised because of non-pure metals,inconsistency when taking potentials and parallax errors when takingthe readings.

Question2

Effectof concentration on cell potential

AddingNH3 to Copper

Cu2+/NH3

Cu2+(aq)+ 2e + NH3(aq) Cu(s)+ NH4+(aq)+ 2e

Cu/Znrxn

Cu(s)+ Zn2+(aq)+2eCu2+(aq)+ Zn(s) + 2e

Inthe initial case, the blue solution changed to colorless and therewas formation of blue solids in the solution. During the reaction,the voltages decreased steadily from +0.928V to +0.685V, the voltagebecome constant after the complete neutralization (Grey, 2014).

Inthe second reaction, the blue copper metal dissolved in the colorlesssolution to make a blue solution. The Voltage changed as per the LeChatelier’s principle where the equation showed a constant readingwhen the solution was completely blue since there was no more metalleft to dissolve in the solution and at that time the solution was atequilibrium.

ii.Adding NH3 to Zinc half cell

Zn2+(aq)+ 2e + NH3(aq) Zn(s)+ NH4+(aq)+ 2e

Cu/Znrxn

Cu(s)+ Zn2+(aq)+2eCu2+(aq)+ Zn(s) + 2e

Inthe reaction between zinc cell and ammonia, the voltage increasedfrom +0.943V to +1.230V as per the Nernst equation when ammonia wasadded to the solution. The equation was balanced after the additionalof Cu ions that oxidized the solution till equilibrium

QuestionThree

Unknown=+ 0.817

Ecell = +0.261

Calculationof Unknown (Ag+) = – 0.817+ 0.261= – 0.556V

Calculationof Ksp, (Grey, 2014)

Usingthe Nernst equation

Ecell=E0-(0.0591/n)logK

0= (-0.556-0.817)-(0.0591/1)log k

1.373=-0.0591logk

Logk = 1.373/-0.0591= –23.07

K= 5.86 x 10-24

Ksp= 5.86×10-24

Identityof the unknow salt is = Iodine with a -0.556V from the experiment

Fromthe table of the literature electrodes, is Iodine which has electrodepotential of -0.54V, which is almost the same with the calculatedvalues.

Partfour

Corrosionof Iron

Theimportance of the phenolphthalein indicator in the experimentdetermined the Ph of the initial solution.

Thepurpose of K3Fe(CN)6

Thesolution added to determine the presence of iron two ions that whenthey react, they form the dark blue compound that a mixture of ironii and iii ions to form the Prussian blue compound (Grey, 2014).

Reference

Grey,A. (2014). .New York, Dream spinner Press.