Could I get some help with these 2 problems. Thanks. 1. A) i.Calculate the value
ID: 1061212 • Letter: C
Question
Could I get some help with these 2 problems. Thanks.
1.
A)
i.Calculate the value of delta G° for the reaction below
Co (s) + Fe^2+ (aq) --> Fe (s) + Co^2+ (aq)
ii. Is the reaction spontaneous as written
B) Calculate the equilibrium constant (Keq) for the reaction below at 25°C.
Co (s) + Fe^2+ (aq) --> Fe (s) + Co^2+ (aq)
2. Calculate the cell potential for this reaction when the concentration of [Cu^2+] = 1.0 × 10^-5 M and [Zn^2+] = 1.0 M.
Zn (s) + Cu^2+ (aq) ---> Cu (s) + Zn^2+ (aq)
Explanation / Answer
A) i)
The given reaction is Co (s) + Fe^2+ (aq) ------------> Fe (s) + Co^2+ (aq)
Let us split the reaction into 2 half reactions
Co (s) ----------> Co ^2+ (aq) + 2e- .............Oxidation (because oxidation number of Co increases from 0 to +2)
Fe ^2+ (aq) 2 e- -----> Fe (s) ...............Reduction ( because oxidation number of Fe decreases from +2 to 0)
If we have an electrochemical cell set up, then oxidation reaction is the one that takes place at the anode , therefore Co will act as an anode & Fe will act as a cathode
From the table of standard reduction potentials, we can get reduction potential values for Fe and Co. The values are
Co^2+ + 2e- -----> Co(s) Eo = -0.28 V
Fe^2+ + 2e- ------> Fe (s) Eo = -0.44 V
Eo cell is calculated as
Eo(cell) = Eo (cathode) - Eo (anode)
we know Cathode is Fe and anode is Co, so we get Eo as
Eo = Eo (Fe) - Eo (Co)
Eo = -0.44 - (-0.28) = -0.44 + 0.28 = -0.26 V
Eo(cell) = -0.26 V
The relationship between Delta Go and Eo is given as
Delta Go = - n F Eo(cell)
where Eo is standard cell potential
n = number of electrons transferred in redox reaction which is 2 mol for the given reaction
F is Faraday's constant = 96485 C/ mol
Delta Go = - 2 mol * 96485 C/ mol * (-0.26 V)
Delta Go = 50172.2 .C.V
1 J is defined also defined as when a charge having 1 coulombs flows through an circuit having potential difference of 1 volt. Therefore 1 J = 1 C.V
using this as conversion factor , we get 50172.2 C. V * 1 J / 1 C.V ( C.V gets cancelled)
So we have Delta Go = 50172.2 J
Delta Go = 50.2 kJ
ii)
The reaction is not spontaneous because value of delta Go is positive. In order to be spontaneous , delta Go should be negative
B)
We will use the value of delta Go that we calculated in previous part
Relationship between Delta Go and equilibrium constant Keq is given as
DeltaGo = - RT ln (Keq)
We know delta Go which is 50172.2 J
T is given as 25 degree C which is 25 + 273 = 298 K
50172.2 J/mol = - 8.314 J/ mol K * 298 K * ln (Keq)
50172.2 J/mol = - 2477.572 J/mol * ln Keq
ln Keq = -20.25................................................J/mol gets cancelled
Keq = e^-20.25
Keq = 1.60 x 10^-9
#2
The reaction is given as Zn (s) + Cu^2+ (aq) --------> Cu (s) + Zn^2+ (aq)
we will first divide the reaction into 2 half reactions
Zn (s) ----------> Zn^2+(aq) +2e- .......................Oxidation
Cu^2+ (aq) +2e- ----------> Cu(s) ......................Reduction
Since Zn is undergoing oxidation, it will act as an anode and Cu will act as a cathode
From the standard reduction potential table, we can get standard electrode potential values for Zn and Cu
For Zn Eo is -0.76 V and
for Cu Eo is 0.337 V
Eo (cell) is calculated as
Eo(cell) = Eo (cathode) - Eo (anode)
Eo(cell) = 0.337 - ( -0.76)
Eo(cell) = 1.097 V
E(cell) is given by Nernst equation
E(cell) = Eo (cell) - 0.0592/n * log [Zn^2+ / Cu^2+]
Here, n is number of e- transferred.
Zn^2+ is 1.0 M
Cu^2+ is 1 x 10^-5 M
substituting these values, we get
E(cell) = 1.097 - 0.0592/2 * log [ 1/1x10^-5]
E(cell) = 1.097 - 0.0296 * log ( 1 x 10^5)
E (cell) = 1.097 - 0.0296*5
E(cell) = 1.097 - 0.148
E(cell) = 0.949 V