Constants Part A Calculate the equilibrium constant for each of the reactions at
ID: 1028331 • Letter: C
Question
Constants Part A Calculate the equilibrium constant for each of the reactions at 25 C Standard Electrode Potentials at 25 °C Reduction Half-Reaction Pb". (aq) + Mg(s) Pb(s) + Mg2+ (aq) Express your answer using three significant figures. E° (V) 0.13 2.37 0.54 1.36 1.21 0.16 Pb"(aq) + 2 e Mg2+ (aq) + 2 e- Pb(s) Mg(s) 12(8) +2 e--+21 (aq) Cl2 (g) + 2 e- 2 Cl-(aq) MnO2 (s) + 4 H , (aq) + 2 e- Cu2+ (aq) +2 e Mn2+ (aq) + 2 H2O(1) Cu(s) Submit t An Part B 12(a) + 2 Cl (aq) 2 1 (aq) + Cl2 (g) Express your answer using two significant figures. Submit Part C MnO2 (s) + 4 H' (aq) + Cu(s) Mn2+ (aq) + 2 H2O(l) + Cu2+ (aq) Express your answer using two significant figures.Explanation / Answer
Pb+2 (aq) + Mg(s) ------> Pb(s) + Mg+2(aq)
here Pb+2 -----> Pb is reduction reaction done at cathode
Mg -----> Mg+2 is oxidation reaction done at anode.
E0 cell = E0 cathode - E0 anode
E0 cell = -0.13 -(-2.37)
E0 cell = 2.24 V
we have dG = -nFE0 cell
where n = number of electrons involved in the reaction
F = faraday = 96500 coloumbs
and also dG = -2.303RTlogK
R = gas constant = 8.314
T = absolute temperature
-nFE0 = -2.303RTlogK
logk = nFE0/2.303RT
logK = (2*96500*2.24)/(2.303*8.314*298)
K = 5.86 * 10^75
I2(s) + 2Cl-(aq) -----> 2I-(aq) + Cl2(g)
here I2 -----> 2I- is reduction reaction done at cathode
2Cl- -----> Cl2 is oxidation reaction done at anode.
E0 cell = E0 cathode - E0 anode
E0 cell = 0.54 -(1.36)
E0 cell = -0.82 V
we have dG = -nFE0 cell
where n = number of electrons involved in the reaction
F = faraday = 96500 coloumbs
and also dG = -2.303RTlogK
R = gas constant = 8.314
T = absolute temperature
-nFE0 = -2.303RTlogK
logk = nFE0/2.303RT
logK = (2*96500*(-0.82))/(2.303*8.314*298)
K = 1.835 * 10^-28
MnO2(s) + 4H+(aq) + Cu(s) -----> Mn+2(aq) + 2H2O(l) + Cu+2(aq)
here Mn+4 -----> Mn+2 is reduction reaction done at cathode
Cu -----> Cu+2 is oxidation reaction done at anode.
E0 cell = E0 cathode - E0 anode
E0 cell = 1.21 -(0.16)
E0 cell = 1.05 V
we have dG = -nFE0 cell
where n = number of electrons involved in the reaction
F = faraday = 96500 coloumbs
and also dG = -2.303RTlogK
R = gas constant = 8.314
T = absolute temperature
-nFE0 = -2.303RTlogK
logk = nFE0/2.303RT
logK = (2*96500*(1.05))/(2.303*8.314*298)
K = 3.2824 * 10^35