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 ΔG_nonstandard 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 E^o_{Pb²⁺ + 2 e^-→Pb} = -0.131 V and E^o_{Sn²⁺ + 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^-)

Answer 1

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