Question

please finish the whole thing, ( Thank you )

a. For a certain chemical reaction, ΔH = - 34.4 kJ and ΔS = - 85.0 J/K

Calculate ΔG or the reaction at 298K

b. By using data in Appendix E in the textbook, determine which is the more unfavorable reduction, Cd2+ (aq) to Cd(s) or Ca2+ (aq) to Ca(s).

c. Using standard reduction potentials (Appendix E in the textbook), calculate the standard emf for each of the following reactions.

2Al3+(aq)+3Ca(s)→2Al(s)+3Ca2+(aq)

d. Using data in Appendix E in the textbook, calculate the standard emf for each of the following reactions.

Cu(s)+Ba2+(aq)→Cu2+(aq)+Ba(s)

2. 3Fe2+(aq)→Fe(s)+2Fe3+(aq)

3. Hg2+2(aq)+2Cu+(aq)→2Hg(l)+2Cu2+(aq)

e. If the equilibrium constant for a one-electron redox reaction at 298 K is 2.9×10⁶, calculate the corresponding ΔG and E cell under standard conditions.

ΔG =

2. E cell =

Answer 1

Calculate ΔG° for the reaction at 298 K.
dG = dH - TdS
dG = - 34400 Joules - (298K)(-85 J/K)

dG = - 9070 Joules
your answer is
dG = - 9.07 kJ

Cd2+(aq) + 2e­ Cd(s) P = -0.403 V
Ca 2+(aq) + 2e­ Ca(s) P = -2.87 V
The reduction of Ca 2+(aq) to Ca(s) is the more energetically unfavorable
reduction because it has a more negative P value.

ΔG°= -RTlnK = -(8.3145)(298)(ln2.9x10^6) = -3.7x10^4 J = -37kJ
to find E°cell, ΔG°=-nFE°cell, which can be manipulated to
E°cell= ΔG°/-nF = (-3.7x10^4)/ (1x96485) = 0.38V