Question

Calculate the non standard Gibbs Free Energy change, ΔG at 250C for the following reaction with the indicated concentrations.

Zn + 2Ag+ (0.30*M*) → 2Ag + Zn2+ (0.50*M*)

Remember you'll need to calculate both the standard and non standard cell potential Ecell.

Remember ΔG0 = − nFE0cell and a similar equation for the NONstandard value ΔG = −nFEcell

Here are the standard reduction potentials:

Zn2+/ Zn: - 0.763V

Ag+/ Ag: + 0.799V

Answer in kJ to 3SF including signs as needed

Part 2

Calculate the value of **lnK** where K is the
equilibrium constant at 250C for the following reaction with the
indicated concentrations.

Zn + 2Ag+ (0.30*M*) → 2Ag + Zn2+ (0.50*M*)

You should have already calculated both the standard and non standard cell potentials.

Remember ΔG0 = − nFE0cell and ΔG0 = - RTlnK

You *must* use the standard value of ΔG to find K!!!

Think about what value of Ecell you need to use!!

Give your answer to **lnK** to 3SF

Answer #1

**Zn + 2Ag+ (0.30M) ----> 2Ag + Zn2+
(0.50M)**

**E0cell = Ecathode - E0anode**

** = 0.799
-(-0.763)**

** = 1.562
v**

**DG0 = -nFE0cell**

** = -2*96500*1.562**

** = -301.5 kj**

** = -3.01*10^2 kj**

**Ecell = E0cell-
(0.0591/n)log([Zn2+]/[Ag+]^2**

** =
1.562-(0.0591/2)log(0.5/0.3^2)**

** = 1.54 v**

**DG = - nFEcell**

** = -2*96500*1.54**

** = -297.22 kj**

** = -2.97*10^2 kj**

**part 2**

**DG0 = -3.01*10^2 kj**

**DG0 = - RTlnK**

**-3.01*10^5 = -8.314*298lnK**

**lnK = 121.5**

**K = 5.79*10^52**

1) The free energy change for the following reaction at 25 °C,
when [Pb2+] = 1.18 M
and [Cd2+] =
7.90×10-3 M, is -65.9
kJ:
Pb2+(1.18 M) +
Cd(s)>
Pb(s) +
Cd2+(7.90×10-3
M) ΔG = -65.9 kJ
What is the cell potential for the reaction as written under these
conditions?
Answer: ___V
Would this reaction be spontaneous in the forward or the reverse
direction?
2) Use the standard reduction potentials located in the 'Tables'
linked above to calculate the standard...

What is the standard free energy change and equilibrium constant
for the following reaction at 25 °C?
2Ag+ (aq) + Fe (s) 2 Ag (s) +
Fe2+ (aq)
Given standard electrode potentials:
Ag+ (aq) + e- Ag (s)
E0 = 0.7996 V
Fe2+ (aq) + 2 e- Fe (s)
E0 = - 0.447 V

The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG∘, using the
following equation:
ΔG∘=−RTlnK
where T is a specified temperature in kelvins (usually
298 K) and R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values....

The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG∘, using the
following equation:
ΔG∘=−RTlnK
where T is a specified temperature in kelvins (usually
298 K) and R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values....

Item 5
The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG, using the
following equation:
ΔG∘=−RTlnK
where T is standard temperature in kelvins and
R is the gas constant.
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values.
Part A
Acetylene,...

1) Given the following thermochemical reaction and thermodynamic
data, find Gibbs Free Energy, ΔG, and determine if the reaction is
spontaneous or non-spontaneous at 25 °C?
N2(g) + 3H2(g) → 2NH3(g) ΔH = -91.8 kJ
ΔS[N2] = 191 J / mol · K, ΔS[H2] = 131 J / mol · K, and ΔS[NH3]
= 193 J / mol · K
a.98.3 kJ; Non-Spontaneous
b.-98.3 kJ; Spontaneous
c.32.7 kJ; Non-Spontaneous
d.ΔG = -32.7 kJ; Spontaneous
2) What is the oxidation number...

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn2 ] = 0.821 M and [Ni2 ] =
0.0200 M. Standard reduction potentials can be found here.
Zn(s) + Ni+2(aq)=Zn+2(aq) + Ni(s)

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn 2 ] = 0.768 M and [Sn2 ] =
0.0200 M. Standard reduction potentials can be found here.
Zn(s)+Sn2+(aq)----->Zn2+(aq)+Sn(s)
Zn+2e- ---->Zn = -0.76
Sn+ + 2e- ----> Sn = -0.14

A. Using given data, calculate the change in
Gibbs free energy for each of the following reactions. In each case
indicate whether the reaction is spontaneous at 298K under standard
conditions.
2H2O2(l)→2H2O(l)+O2(g)
Gibbs free energy for H2O2(l) is -120.4kJ/mol
Gibbs free energy for H2O(l) is -237.13kJ/mol
B. A certain reaction has ΔH∘ = + 35.4
kJ and ΔS∘ = 85.0 J/K . Calculate ΔG∘ for the
reaction at 298 K. Is the reaction spontaneous at
298K under standard
conditions?

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn2 ] = 0.837 M and [Fe2 ] =
0.0100 M. Standard reduction potentials can be found here.
Zn(s)+Fe^2+(aq) <--->Zn^2+(aq)+Fe(s)
E= _______ V

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