Balancing Redox Reactions

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Balancing Redox Reactions MnO 4 - Mn 2+ manganese is balanced MnO 4 - Mn 2+ + balance oxygen using H 2 O 4H 2 O balance hydrogen using H + MnO 4 - + Mn 2+ + 4 H 2 O 8H + balance charge using e - MnO 4 - + 8H + Mn 2+ + 4 H 2 O +5e -

description

Balancing Redox Reactions. MnO 4 -  Mn 2+. manganese is balanced. balance oxygen using H 2 O. MnO 4 -  Mn 2+ +. 4 H 2 O. balance hydrogen using H +. MnO 4 - +  Mn 2+ + 4 H 2 O. 8H +. balance charge using e -. MnO 4 - + 8H +  Mn 2+ + 4 H 2 O. + 5e -. - PowerPoint PPT Presentation

Transcript of Balancing Redox Reactions

Page 1: Balancing Redox Reactions

Balancing Redox Reactions

MnO4- Mn2+

manganese is balanced

MnO4- Mn2+ +

balance oxygen using H2O

4H2Obalance hydrogen using H+

MnO4- + Mn2+ + 4 H2O8H+

balance charge using e-

MnO4- + 8H+ Mn2+ + 4 H2O+5e-

Page 2: Balancing Redox Reactions

Oxidation States

MnO4- + 8H+ Mn2+ + 4 H2O+5e-

MnO4-

Oxidation state of oxygen =

Oxidation state of hydrogen =

4 x (-2)+ 1(Mn) = -1 Mn = 7+Mn2+ Mn = 2+

Mn7+ + 5 e- Mn2+

0 in O2

2- in compounds

1+ in compounds0 in H2

Page 3: Balancing Redox Reactions

2H+ + 2e- H2 o = 0.00 VMnO4

- + 8H+ Mn2+ + 4 H2O+5e- o = 1.51 V

reduction reaction

MnO4-+8H++5e-Mn2++4 H2O

oxidation reaction

H2 2H++2e-

anode

Pt(s)

2MnO4- + 16H+ +10e- + 5H2 2Mn2+ + 8 H2O + 10H+ + 10e-

2MnO4- + 6H+ + 5H2 2Mn2+ + 8 H2O

cathode

H2(g)(1atm)H+(aq)(1M) Pt(s) Mn2+(aq)(1M)MnO4-(aq)(1M),

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,H+(1M)

( )5 ( )2

Page 4: Balancing Redox Reactions

2H+ + 2e- H2 o = 0.00 V

MnO4- + 8H+ Mn2+ + 4 H2O+5e- o = 1.51 V

ocell

=

Go = -nFo

2MnO4- + 6H+ + 5H2 2Mn2+ + 8 H2O

Go =

ored - o

ox = 1.51 - 0 = 1.51 V

= - (10)(96,500)(1.51) = -1457 kJ

-RT ln K= -1457 kJ

-RT -RTK = e588

Page 5: Balancing Redox Reactions

Nernst Equation

G = -nFcell Go = -nFocell

G =

cell = ocell - RT

nFln Q

standardnon-standard

Go + RT ln Q

Page 6: Balancing Redox Reactions

Nernst Equation

Cu(s)

cell = ocell - RT ln Q

nF

ocell = 0.80 - 0.34 = 0.46 V

Cu2+ + 2e- Cu Ag+ + e- Ag(s)

ored - o

ox =

o= 0.34 V o= 0.80 V

2( )b) reductionoxidation

anode cathode

Cu2+(aq) (1M )4M

Cu (s) Cu2+ + 2e-

a) oxidationAg+(aq) (1M) Ag(s)

Page 7: Balancing Redox Reactions

Nernst Equation

Cu(s) Cu2+(aq) (4M) Ag+(aq) (1M) Ag(s)

cell = ocell - RT ln Q

nF

Cu (s) Cu2+ + 2e- 2Ag+ + 2e- 2Ag (s)

Q = = [Cu2+][Ag+]

= 4 = 412

[products]minitial

[reactants]ninitial

anode cathode

2

Page 8: Balancing Redox Reactions

Nernst Equation

Cu(s) Cu2+(aq) (4M) Ag+(aq) (1M) Ag(s)

cell = ocell - RT ln Q

nF

Cu(s) Cu2+ + 2e- 2Ag+ + 2e- 2Ag

cell = 0.46V 2

= 0.44 V- (298) ln 4(96,500)

(8.314)

anode cathode

Page 9: Balancing Redox Reactions

Non-standard conditions

G =

= -85 kJ

-(2 mol e-) (96,500 C/mol e-)(0.44V)

Cu(s) Cu2+(aq) (4M) Ag+(aq) (1M) Ag(s)

Cu(s) Cu2+ + 2e- 2Ag+ + 2e- 2Ag

Go =

= -89 kJ

-(2 mol e-) (96,500 C/mol e-)(0.46V)

G = -nF

Page 10: Balancing Redox Reactions

Concentration cellCu(s)

Cu(s) Cu2+

oxidation reduction

cell =

ocell = o

red - oox = 0.34 - 0.34 = 0

ocell - RT

nFln Q

Cu2+(aq) (0.5M) Cu2+(aq) (2M) Cu(s)

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

0 V

Page 11: Balancing Redox Reactions

Concentration CellsCu(s) Cu2+(aq) (0.5M)Cu2+(aq) (2M) Cu(s)

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

= [products]minitial

= ocell

= [Cu2+]

[reactants]n

initialanode

cathode

= 0.5 = 0.25 2

Q

cell = -RT 2F

ln .25 =.02 V

oxidation reduction

[Cu2+]cathode

-RT nF

ln Q 0

Page 12: Balancing Redox Reactions

Concentration Cells

P680

QA

+-h

Mn 2H2O 4H+

pH = 3.0

pH = 7.0 cell = ocell - RT ln Q

nFcell = o

cell - 0.059 log Q n

ADP + Pi ATP G > 0

Q = 10-7 / 10-3 cell = 0.24 V G < 0

+ O2 + 4e-+