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© 2006 Brooks/Cole - Thomson

Balancing Equations for Redox Reactions

Some redox reactions have equations that must be balanced by special techniques.

MnO4- + 5 Fe2+ + 8 H+---> Mn2+ + 5 Fe3+ + 4 H2O

Mn = +7 Fe = +2 Fe = +3Mn = +2

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Reduction of VO2+ with Zn

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Balancing Equations

Balance the following in acid solution—

VO2+ + Zn ---> VO2+ + Zn2+

Step 1:Write the half-reactionsOx Zn ---> Zn2+

Red VO2+ ---> VO2+

Step 2:Balance each half-reaction for mass.Ox Zn ---> Zn2+

Red VO2+ ---> VO2+ + H2O2 H+ +

Add H2O on O-deficient side and add H+ on other side for H-balance.

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Balancing Equations

Step 3:Balance half-reactions for charge.Ox Zn ---> Zn2+ + 2e-

Red e- + 2 H+ + VO2+ ---> VO2+ + H2O

Step 4:Multiply by an appropriate factor.Ox Zn ---> Zn2+ + 2e-

Red 2e- + 4 H+ + 2 VO2+

---> 2 VO2+ + 2 H2O

Step 5:Add balanced half-reactions

Zn + 4 H+ + 2 VO2+

---> Zn2+ + 2 VO2+ + 2 H2O

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Tips on Balancing Equations

• Never add O2, O atoms, or O2- to balance oxygen ONLY add H2O or OH-.

• Never add H2 or H atoms to balance hydrogen ONLY add H+

or H2O.• Be sure to write the correct

charges on all the ions.• Check your work at the end to

make sure mass and charge are balanced.

• PRACTICE!

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Using Standard Potentials, Eo

• In which direction do the following reactions

go?

• Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)

–Goes right as written

• 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s)

–Goes LEFT opposite to direction written

• What is Eonet for the overall reaction?

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Cd --> Cd2+ + 2e-or

Cd2+ + 2e- --> Cd

Fe --> Fe2+ + 2e-or

Fe2+ + 2e- --> Fe

Eo for a Voltaic Cell

All ingredients are present. Which way does reaction proceed? Calculate Eo for this cell.

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E at Nonstandard Conditions

• The NERNST EQUATION• E = potential under nonstandard conditions

• n = no. of electrons exchanged

• F = Faraday’s constant

• R = gas constant

• T = temp in Kelvins

• ln = “natural log”

• Q = reaction quotient

QnF

RTcello

cell EE ln

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Eo and Thermodynamics

• Eo is related to ∆Go, the free energy change for the reaction.

• ∆G˚ is proportional to –nE˚

∆Go = -nFEo where F = Faraday constant

= 9.6485 x 104 J/V•mol of e-

(or 9.6485 x 104 coulombs/mol)and n is the number of moles of electrons

transferred

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Eo and ∆Go

∆Go = - n F Eo For a product-favored reaction Reactants ----> Products

∆Go < 0 and so Eo > 0Eo is positive

For a reactant-favored reaction Reactants <---- Products

∆Go > 0 and so Eo < 0Eo is negative

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Eo and Equilibrium Constant

DGo = -RT ln K

DGo = -nFEo

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ElectrolysisUsing electrical energy to produce chemical change.

Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)

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ElectrolysisElectric Energy ---> Chemical Change

•  Electrolysis of

molten NaCl.

•  Here a battery

“pumps” electrons

from Cl- to Na+.

•  NOTE: Polarity of

electrodes is

reversed from

batteries.

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Electrolysis of Molten NaCl

Figure 20.18

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Electrolysis of Molten NaCl

Anode (+)

2 Cl- ---> Cl2(g) + 2e-

Cathode (-)

Na+ + e- ---> Na

Eo for cell (in water) = - 4.07 V (in water)

External energy needed because Eo is (-).

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Electrolysis of Aqueous NaI

Anode (+): 2 I- ---> I2(g) + 2e-Cathode (-): 2 H2O + 2e- ---> H2 + 2 OH-

Eo for cell = -1.36 V

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© 2006 Brooks/Cole - Thomson

Electrolysis of Aqueous CuCl2

Anode (+)

2 Cl- ---> Cl2(g) + 2e-

Cathode (-)

Cu2+ + 2e- ---> Cu

Eo for cell = -1.02 V

Note that Cu2+ is more

easily reduced than

either H2O or Na+.

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Michael Faraday1791-1867

Originated the terms anode, cathode, anion, cation, electrode.

Discoverer of • electrolysis• magnetic props. of matter• electromagnetic induction• benzene and other organic

chemicalsWas a popular lecturer.

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Quantitative Aspects of Electrochemistry

Consider electrolysis of aqueous silver ion.Ag+ (aq) + e- ---> Ag(s)1 mol e- ---> 1 mol AgIf we could measure the moles of e-, we

could know the quantity of Ag formed.But how to measure moles of e-?

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But how is charge related to moles of electrons?

Quantitative Aspects of Electrochemistry

= 96,500 C/mol e- = 1 Faraday

Michael Faraday1791-1867

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Quantitative Aspects of Electrochemistry

1.50 amps flow through a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited?

Solution(a) Calc. charge

Charge (C) = current (A) x time (t)= (1.5 amps)(15.0 min)(60 s/min) = 1350 C

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© 2006 Brooks/Cole - Thomson

Quantitative Aspects of Electrochemistry

Solution(a) Charge = 1350 C(b) Calculate moles of e- used

1.50 amps flow through a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited?

(c) Calc. quantity of Ag

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© 2006 Brooks/Cole - Thomson

Quantitative Aspects of Electrochemistry

The anode reaction in a lead storage battery is

Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?

Solutiona) 454 g Pb = 2.19 mol Pbb) Calculate moles of e-

c) Calculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C

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© 2006 Brooks/Cole - Thomson

Quantitative Aspects of Electrochemistry

The anode reaction in a lead storage battery is

Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?

Solutiona)454 g Pb = 2.19 mol Pbb) Mol of e- = 4.38 molc)Charge = 423,000 C

About 78 hours

d) Calculate time