Chapter 21 Lecture 5

Chapter 21 Lecture 5 - Using the Nernst Equation to...

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Using the Nernst Equation to Calculate E cell SAMPLE PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn 2+ half-cell and an H 2 /H + half-cell under the llowing conditions: following conditions: [Zn 2+ ] = 0.010 M [H + ] = 2.5 M P = 0.30 atm H 2 Calculate E cell at 298 K. PLAN: SOLUTION: Find E o cell and Q in order to use the Nernst equation. Determining E o cell : E o = 0.00 V 2H + ( aq ) + 2e - H 2 ( g ) [Zn 2+ E o = -0.76 V Zn 2+ ( aq ) + 2e - Zn( s ) Zn( s ) Zn 2+ ( aq ) + 2e - E o cell = +0.76 V Q = P x [Zn ] H 2 [H + ] 2 .30)(0.010) Q = 4.8x10 -4 Q = (0.30)(0.010) (2.5) 2 E ell = E o ell - 0.0592 V g Overall: Zn( s ) + 2H + ( aq ) Zn 2+ ( aq ) + H 2 ( g ) 21-55 cell cell n log Q E cell = 0.76 - (0.0592/2)log(4.8x10 -4 ) = 0.86 V
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Changes in Potential During Cell Operation Zn( s ) + Cu 2+ ( aq ) Zn 2+ ( aq ) + Cu( s ) ( )C u ) u ( E cell = E o cell -l n Q RT nF When Q is large enough, E ell =0 equilibrium cell No more free energy is released so the cell can do ln Q nF no more work The battery is dead 21-56
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Concentration Cell contains the same redox couple in both the anode and cathode half- cells: E o cell = E o cathode –E o anode = 0 E cell = E o cell -= - ln Q RT nF ln Q nF Overall (cell) reaction Cu 2+ ( aq ,1.0 M) u 2+ q 01M) Cu ( aq , 0.1 M) Oxidation half-reaction Cu( s ) Cu 2+ ( aq , 0.1 M) + 2e - Reduction half-reaction Cu 2+ ( aq , 1.0 M) + 2e - Cu( s ) 21-57
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Applications of Concentration Cell The laboratory measurement of pH eference Pt Reference (calomel) electrode Glass electrode Paste of Hg 2 Cl 2 in Hg AgCl on Ag on Pt Cl Hg 1 M HCl hin glass orous ceramic KCl solution 21-58 Thin glass membrane Porous ceramic plugs
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Calculating the Potential of a Concentration Cell SAMPLE PROBLEM: A concentration cell consists of two Ag/Ag + half-cells. In half-cell A, electrode A dips into 0.0100 M AgNO 3 ; in half-cell B, electrode B dips into 4.0x10 -4 M AgNO . What is the cell potential at 298 K? 3 Which electrode has a positive charge? PLAN: E o cell will be zero since the half-cell potentials are equal. E cell is calculated from the Nernst equation with half-cell A (higher [Ag + ]) having Ag + being reduced and plating out, and in half-cell B Ag( s )
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This note was uploaded on 07/31/2010 for the course CHEM CHEM 01C taught by Professor Unknown during the Spring '10 term at UC Riverside.

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Chapter 21 Lecture 5 - Using the Nernst Equation to...

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