Calculate the cell potential, at 25 C, based upon the overall reaction
Zn2+(aq) + 2 Fe2+(aq) -> Zn(s) + 2 Fe3+(aq)
if [Zn2+] = 1.50 x 10-4 M, [Fe3+] = 0.0200 M, and [Fe2+] = 0.0100 M. The standard reduction potentials are as follows:
Zn2+(aq) + 2 e– --> Zn(s) E = –0.763 V
Fe3+(aq) + e– --> Fe2+(aq) E = +0.771 V
a. –1.665 V
b. –1.534 V
c. –1.439 V
d. –0.008 V
e. +0.008 V ----> Answer
Which of the following statements is/are CORRECT?
1. Naturally occurring isotopes account for only a small fraction of known radioactive isotopes.
2. A few radioactive isotopes with long half-lives, such as U-235 and U-238, are found in nature.
3. Trace quantities of some short-lived radioactive isotopes, such as C-14, are found in nature because they are formed continuously by nuclear reactions.
a. 1 only
b. 2 only
c. 3 only
d. 1 and 2
e. 1, 2, and 3 ----> Answer
I don't agree with a
I agree with b although I'm not sure of "what a small fraction" is. .
To calculate the cell potential based on the given reaction and concentrations, we can use the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (25°C = 298 K)
n is the number of electrons transferred in the balanced equation (2 in this case)
F is Faraday's constant (96,485 C/mol)
ln is the natural logarithm
Q is the reaction quotient
First, let's calculate the reaction quotient, Q:
Q = ([Zn]^1 * [Fe3]^2) / ([Zn2]^1 * [Fe2]^2)
Plugging in the given concentrations:
Q = ((1.50 x 10^(-4))^1 * (0.0200)^2) / ((0.0100)^2)
Q = 6.000 x 10^(-7) / 10^(-4)
Q = 6.000 x 10^(-3)
Now, let's calculate the cell potential:
Ecell = (-0.763 V) - ( (8.314 J/(mol·K)) * (298 K) / (2 * 96,485 C/mol) ) * ln(6.000 x 10^(-3))
Ecell = -0.763 V - (0.0257 V) * ln(6.000 x 10^(-3))
Using a calculator, we find:
Ecell = -1.534 V
Therefore, the correct answer is (b) -1.534 V.
For the second question, the correct answer is (e) 1, 2, and 3.
To calculate the cell potential of the given reaction, we need to use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- (0.0592/n) is the conversion factor at 25°C (n is the number of electrons transferred)
- log(Q) is the logarithm of the reaction quotient
Let's break down the given reaction to identify the number of electrons transferred.
Zn2+(aq) + 2 Fe2+(aq) -> Zn(s) + 2 Fe3+(aq)
We see that 2 electrons are transferred per mole of reaction.
Now, let's calculate the reaction quotient Q using the given concentrations:
Q = [Zn(s)] * [Fe3+]^2 / [Zn2+] * [Fe2+]^2
Q = (1) * (0.0200)^2 / (1.50 x 10^-4) * (0.0100)^2
Next, we need to substitute the values into the Nernst equation to find the cell potential:
Ecell = -0.763 V - (0.0592/2) * log((0.0200)^2 / (1.50 x 10^-4) * (0.0100)^2)
Calculating this gives us Ecell = -1.439 V.
Therefore, the correct answer is option c: –1.439 V.
For the second question, let's analyze each statement:
1. Naturally occurring isotopes account for only a small fraction of known radioactive isotopes.
This statement is true because most radioactive isotopes are not naturally occurring.
2. A few radioactive isotopes with long half-lives, such as U-235 and U-238, are found in nature.
This statement is true because uranium isotopes U-235 and U-238 have long half-lives and occur in nature.
3. Trace quantities of some short-lived radioactive isotopes, such as C-14, are found in nature because they are formed continuously by nuclear reactions.
This statement is true because short-lived radioactive isotopes like carbon-14 (C-14) are constantly formed by nuclear reactions and can be found in trace amounts in nature.
Considering all three statements are correct, the correct answer is option e: 1, 2, and 3.