Zn Zn 2 2 e Cl 2 2 e 2 Cl Zn Cl 2 ZnCl 2 Therefore n 2 369000 J 2 96500 E Thus

Zn zn 2 2 e cl 2 2 e 2 cl zn cl 2 zncl 2 therefore n

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Zn Zn 2+ + 2 e Cl 2 + 2 e 2 Cl Zn + Cl 2 ZnCl 2 Therefore… n = 2 369,000 J = (2) (96500) ( E 0 ) Thus… E 0 = ) 96500 )( 2 ( 369000 = 1.91 V
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19-17 Thermodynamics, Equilibrium and Electrochemistry: A Summary G 0 = nF E 0 also G 0 = RT ln (K) Therefore… nF E 0 = RT ln(K) ln (K) = nF E 0 / RT At 25 0 C, F RT = 0.025693 J/C = 0.025693 V Therefore… ln (K) = V 0.025693 nE 0 or.. E 0 = n 0.025693 ln (K)= n 0.0592 log (K) e.g. Calculate the equilibrium constant at 25 0 C for the reaction… Cr 2 O 7 2 + 6 Br + 14 H + 2 Cr 3+ + 3 Br 2 + 7 H 2 O Given: 3 2 Br Br 2 + 2 e E 0 = 1.09 V 1 Cr 2 O 7 2 + 14 H + + 6 e 2 Cr 3+ + 7 H 2 O E 0 = +1.33 V Cr 2 O 7 2 + 6 Br + 14 H + 2 Cr 3+ + 3 Br 2 + 7 H 2 O E 0 = +0.24 V NOTE: a) balancing the equation is necessary to find (n), the number of electrons transferred. b) E 0 is an intensive property. Its value does not change when the half reactions are multiplied by small integers to balance the number of electrons. Back to our equation… ln (K) = V 0.025693 nE 0 = V 0.025693 (6)(0.24) = 56 K = e 56 = 2.1 x 10 24 E 0 is positive therefore K is greater than 1.
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19-18 Spontaneity of Redox Reactions Under Conditions Other Than Standard State Consider the reaction: Zn(s) + Cu 2+ ( aq,1M ) Zn 2+ ( aq,1M) + Cu(s) E 0 cell = 1.103 V What happens when one increases the concentration of the Cu 2+ ion or decreases the concentration of the Zn 2+ (ion)? ( nonstandard states)Answer: the equilibrium will shift to the right i.e. the reaction under these concentrations is even more spontaneous E cell > E 0 cell . Zn(s) + Cu 2+ ( aq,1.5M ) Zn 2+ ( aq,0.75M) + Cu(s) E 0 cell = 1.142 V The Nernst Equation Non-standard states G = G 0 + RT ln Q nF E = nF E 0 + RT ln Q by dividing both sides by –nF, one obtains… E cell = E 0 cell nF RT ln Q at 25 0 C R,T, and F are constants F RT = 0.02568 E cell = E 0 cell n 0.02568 ln Q Q is the reaction quotient. NOTE: E cell = E 0 cell n 0.02568 ln Q or E cell = E 0 cell - n 0592 . 0 log Q 1. When Q = K (i.e. equilibrium is reached), then E cell = 0. The components in the two compartments of the cell have the same free energy. Thus, G = 0. There is no driving force, the battery is drained (dead) 2. When we have standard conditions (concentration fo ions = 1M, pressure = 1atm) then Q = 1 and ln Q = 0. Thus, E cell = E 0 cell . 3. When Q > 1 (the concentration of product more than those of reactants), then E cell < E 0 cell (check the equation above). 4. When Q < 1 (the concentration of product less that those of reactants), then E cell > E 0 cell (check the equation above).
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19-19 The relationship between E 0 and K therefore is… G 0 = nF E 0 G 0 = RT ln K nF E 0 = RT ln K E 0 cell = nF RT ln K R, F are constants, T = 25 0 C E 0 cell = n 0.02568 ln K Does 0.02568 have units? What is it? It is (V) e.g. Consider the reaction… Cu 2+ (aq) + Zn (s) Zn 2+ (aq) + Cu (s) E 0 cell = 1.10 V (calculated from the E 0 cathode - E 0 anode ) Calculate the cell potential when [Cu 2+ ] = 3.0 M and [Zn 2+ ] = 0.10 M E cell = E 0 cell n 0.02568 ln ] [Cu ] [Zn 2 2 (where, n = 2) E cell = 1.10 2 0.02568 ln 0 . 3 0.10 = 1.10 + 0.04 = 1.14 V In other words, the reaction is even more spontaneous under these non-standard conditions, and this is consistent with Le Chatelier’s Principle (increasing [Cu 2+ ] and/or decreasing [Zn 2+ ] will shift the reaction to the right.
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