© university of south carolina board of trustees oxidation states example find the oxidation state...
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Oxidation States Example
Find the oxidation state of…
Fe: Fe(s) + O2 ® Fe2O3
Al: Al(s) + O2 ® AlO2
Can this compound exist?
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Chapt 18Electrochemistry
Sec. 3Voltaic Cells: Experimental
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Voltaic Cells
A cell with a spontaneous flow of current
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Salt Bridge RequiredTwo Reaction Compartments
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Direction of electron flow?
Cu(s) Cu2+ + 2e- (oxid.)
Cu2+ + 2e- Cu(s) (reduc.)
Ag+ + e- Ag(s) (reduc.)
Ag(s) Ag+ + e- (oxid.)
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Cu(s) Cu2+ + 2e- (oxid.)
Cu2+ + 2e- Cu(s) (reduc.)
Ag+ + e- Ag(s) (reduc.)
Ag(s) Ag+ + e- (oxid.)
Positive Voltage, Spontaneous electron flow
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Half Cell
Sn2+ and Sn4+ are both in contact with an inert electrode
Inert: ● electrical conductor● no chemical
reactions● e.g., Pt, graphite
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Hydrogen gas is bubbled over an inert platinum electrode
Half Cell: Gas Reactants
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The Calomel Electrode
Hg2Cl2(s) (calomel) is reduced to Hg() in contact with a Pt wire.
Common standard for comparison to other electrodes.
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No Salt BridgeOne Reaction Compartment
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Batteries are Solid Voltaic Cells
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Chapt 18Electrochemistry
Sec. 4Voltaic Cells: Determining Cell Voltage
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Half-Cell Voltages
2H+(aq) + 2e- H2(g) E° = 0.00 V
by definition
A voltage can be assigned to a ½ reaction, if we establish a reference point
Sum of ½-Cell Voltages
=Voltage of the Cell
Voltage or Potential
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Standard Potentials
Value of voltage of ½-reaction is versus H2 ½-reaction
Standard StatesLiquid / Solid = pure
Gas = 1 atm
Solution = 1 M
Written as a reductionReverse sign, if reaction changes direction
Do not multiply by stoichiometry factor
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Activity Series
+2.87 V
+0.77 V
+0.34 V
+0.00 V
-0.25 V
-1.66 V
-3.05 V
High Reduction PotentialEasy to Reduce
Low Reduction PotentialHard to Reduce
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Potentials and Spontaneity
A positive cell potential is spontaneous
Half-cell with the larger (positive) potential goes forward (reduction).
Half-cell with the lower potential (small positive or large negative) goes in reverse (oxidation).
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Example: Spontaneity
Determine the spontaneous direction for the following cell:
Pb+2(aq) + 2e− Pb(s)
E° = −0.126 V
Br2() + 2e− 2Br−(aq)
E° = +1.060 V
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Putting together a voltaic cellA voltaic cell is made up of an Al electrode in a solution of Al(NO3)3 and a Cu electrode in a solution of Cu(NO3)2 with a potassium nitrate salt bridge.
Al3+(aq) + 3e− Al(s) E° = −1.66 V
Cu2+(aq) + 2e− Cu(s) E° = +0.34 V
What is the cell potential?Which electrode is positive?Which electrode is negative?What is the oxidizing agent?What is the reducing agent?
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Student Example
Write the cell reaction and calculate E°cell for the voltaic cell made up of a standard Mg/Mg2+ half-cell and a standard Fe2+/Fe3+ half-cell.
Mg2+(aq) + 2e− Mg(s) E° = -2.37 V
Fe3+(aq) + e− Fe+2
(aq) E° = +0.77 V
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Activity Series
+2.87 V
+0.77 V
+0.34 V
+0.00 V
-0.25 V
-1.66 V
-3.05 V
High Reduction PotentialEasy to Reduce
Low Reduction PotentialHard to Reduce
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Activity Series
+2.87 V
+0.77 V
+0.34 V
+0.00 V
-0.25 V
-1.66 V
-3.05 V
High Reduction PotentialEasy to Reduce
Low Reduction PotentialHard to Reduce
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Activity SeriesHigh Reduction Potential
Easy to Reduce Hard to Oxidize
Low Reduction PotentialHard to Reduce Easy to Oxidize
+2.87 V
+0.77 V
+0.34 V
+0.00 V
-0.25 V
-1.66 V
-3.05 V
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Example: Activity Series
Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Co(NO3)2?
Co2+(aq) + 2e- Co(s) E° = -0.28 V
Fe2+(aq) + 2e- Fe(s) E° = -0.44 V
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Example: Activity Series
Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Co(NO3)2?
Co2+(aq) + 2e- Co(s) E° = -0.28 V
Fe2+(aq) + 2e- Fe(s) E° = -0.44 V
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Example: Activity Series
Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Zn(NO3)2?
Zn2+(aq) + 2e- Zn(s) E° = -0.76 V
Fe2+(aq) + 2e- Fe(s) E° = -0.44 V
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Example: Activity Series
Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Zn(NO3)2?
Zn2+(aq) + 2e- Zn(s) E° = -0.76 V
Fe2+(aq) + 2e- Fe(s) E° = -0.44 V