Lecture31 - Chem 162 Lecture 31 Spring 2011 Calculating...

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Unformatted text preview: Chem 162, Lecture 31, Spring 2011 Calculating Cell Voltage under Non-standard Conditions ∆ G = ∆ G o + RT ln Q -nFE = -nFE o + RT ln Q E = E o- RT nF ln Q Nernst Equation Converting to base 10 logarithms: E = E o – (RT/nF)ln Q = E o- 2.303RT nF log Q Substituting: T = 298.15 K F = 96,485 C/mol R = 8.314 Joule/K RT/F = 0.025693 V 2.303RT F = (2.303)(8.314)(298) 96485 = 0.059161 V E = E o- 0.0591 n log Q Nernst Equation 0.0592 actually! 1 Example: 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 cathode Al 3+ (aq) + 3e- → Al(s) -1.68 anode E o = E(cathode) – E(anode) = – 1.18 V – (-1.68 V) = + 0.50 V Using the Nernst equation: E = E o- 0.0591 n log Q E = 0.50 - 0.0591 6 log (1.50) 2 (0.010) 3 = 0.44 volts Another Example: Cu + 2Ag + → 2Ag + Cu 2+ E o = 0.46 volts At equilibrium: E = 0 and Q = K 0 = E o- 0.0591 n log K => log K = n E o 0.0591 => log K = 2(0.46) 0.0591 = 15.567 => K = 10 15.567 = 3.7 x 10 15 2 Starting with standard conditions, what is the voltage of the cell after 99% has reacted?...
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This note was uploaded on 11/09/2011 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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Lecture31 - Chem 162 Lecture 31 Spring 2011 Calculating...

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