1) Consider the reaction: A (aq) B (aq) at 253 K where the initial concentration
ID: 957994 • Letter: 1
Question
1) Consider the reaction:
A (aq) B (aq)
at 253 K where the initial concentration of A = 1.00 M and the initial concentration of B = 0.000 M. At equilibrium it is found that the concentration of B = 0.523 M. What is the maximum amount of work that can be done by this system when the concentration of A = 0.859 M?
2) Consider the following reaction at 257 K:
2 A (aq) + 1 B (aq) 2 C (aq) + 2 D (aq)
An experiment was performed with the following intitial concentrations: [A]i = 1.37 M, [B]i = 2.19 M, [C]i = 0.47 M, [D]i = 0.31 M. The reaction was allowed to proceed until equilibrium was reached at which time it was determined that [A] = 0.57 M. What is Gnonstandard at the initial conditions? (I got -14.16kJ but it said i was wrong)
3) Consider the cell described below at 287 K:
Sn | Sn2+ (1.27 M) || Pb2+ (2.49 M) | Pb
Given EoPb2+ + 2 e-Pb = -0.131 V and EoSn2+ + 2 e-Sn = -0.143 V. What will the concentration of the Pb2+ solution be when the cell is dead?
4)A concentration cell is built based on the reaction:
2 H+ + 2 e- H2
The pH in one of the half cells is -0.03743, while the pH in the other is 2.804. If the temperature of the overall cell is 296 K, what is the potential? (Faraday's constant is 96,485 C/mol e-)
Explanation / Answer
3) Sn | Sn2+ (1.27 M) || Pb2+ (2.49 M) | Pb
Pb2+ + 2 e-Pb = -0.131 V (reduction )
Sn2+ + 2 e-Sn = -0.143 V (oxidation )
net reaction
Pb2+ + Sn ------> Pb + Sn2+
the concentration of the Pb2+ solution be when the cell is dead it means that Ecell = 0
the E0cell = E0reduction - E0oxidation = -0.131 V - (-0.143 V) = 0.012 V
hence
Ecell = E0cell - RT/nF ln Sn2+ / Pb2+
0 = 0.012 - 8.314 * 287 * 2.303 / 2* 96500 log 1.27 / PB2+
-0.012 = - 0.02847 log 1.27 /PB2+
0.012 / 0.02847 = -log PB2+ /1.27
(anti -0.42149 ) = PB2+ /1.27
0.3788 = PB2+ /1.27
PB2+ = 0.4810 M
4) let us consider
2 H+ (C1) + 2 e- ----->H2 pH 1= -0.03743
2 H+ (C2) + 2 e- ----->H2 pH 1 = 2.804
hence the net reaction is given by
H+ (C2) ----> H+ (C1)
Ecell = - RT /nF lln C1/C2 = RT /nF lnC2/C1
= 2.303 * 8.314 * 296 / 2* 96485 log C2 /C1
multiplying by - to convert into pH
= 0.02937 * log pH 2/pH1
= 0.02937* log 2.804 / 0.03743
= 0.02937 (- log 74.91)
= 0.02937 (- 1.872)= 0.05498 V
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