Asked by JuYoung
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
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
Answers
Answered by
DrBob222
I don't agree with a
I agree with b although I'm not sure of "what a small fraction" is. .
I agree with b although I'm not sure of "what a small fraction" is. .
Answered by
Autumn
Hey, for the first one, you would solve using equation Ecell=Ecathode-Eanode. You can see in the forumula that the Zinc half equation is the cathode because Reduction (electrons gained) happens at the cathode. Thus your equation becomes Ecell=-.763-.771. Plugging this into a calculator, we get -1.534, or B.
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