Download - Balancing Equations for Redox Reactions
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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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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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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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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