lecture+20sfa

# lecture+20sfa - Calculation of Voltage under Non-standard...

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Calculation of Voltage under Non-standard Conditions G = G o + RT ln Q -nFE = -nFE o + RT ln Q E = E o - nF RT ln Q Nernst Equation Converting to base 10 logarithms: E = E o - log Q Substituting: T = 298 K F = 96485 R = 8.314 Joule = = 0.0592 E = E o n 0592 . 0 log Q Nernst Equation 2Al(s) + 3Mn 2+ (aq) → 2Al 3+ (aq) + 3Mn(s) What is the voltage of such a cell if: [Mn 2+ ] = 0.010 M [Al 3+ ] = 1.50 M E o (volts) Mn 2+ (aq) + 2e - → Mn(s) -1.18 Al 3+ (aq) + 3e - → Al(s) -1.66 E o = 0.48 volts for this cell. Using the Nernst equation: E = E o n 0592 . 0 log Q

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E = 0.48 - log 3 2 ) 010 . 0 ( ) 50 . 1 ( E = 0.42 volts Cu + 2Ag + → 2Ag + Cu 2+ E o = 0.46 volts Starting with standard conditions, what is the voltage of the cell after 99% has reacted? Cu(s) + 2Ag + → 2Ag(s) + Cu 2+ E o = 0.46 volts Start 1.0 1.0 Change -0.99 +0.495 End 0.01 1.495 (not at equilibrium yet) Q = 2 2 ] Ag [ ] Cu [ + + = 2 01 . 0 495 . 1 = 14,950 E = E o n 0592 . 0 log Q E = 0.46 - 2 0592 . 0 log 14,950 = 0.34 volts What is the cell voltage at equilibrium? At equilibrium: E = 0 and Q = K
E = E o n 0592 . 0 log Q At equilibrium: 0 = E o - n 0592 . 0 log K 0592 . 0 nE K log o = What is K for: Cu + 2Ag + → 2Ag + Cu 2+ log K = 0592 . 0 ) 46 . 0 ( 2 = 15.567 K = 10 15.567 = 3.7 x 10 15 Alternate approach using natural logs: E = E o - nF RT ln Q At 298 K: F RT = 485 , 96 ) 15 . 298 ( 3145 . 8 = n 025693 . 0 E = E o - n 025693 . 0 ln Q At equilibrium: E = 0 and Q = K 0 = E o - n 025693 . 0 ln K ln K = 025693 . 0 nE o For the above problem: ln K = 025693 . 0 ) 46 (. 2 = 35.807

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35.807 = 3.6 x 10 15 What is K for: 2Ag + Cu 2+ → Cu + 2Ag + log K = 0592 . 0 ) 46
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lecture+20sfa - Calculation of Voltage under Non-standard...

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