What is the cell potential for the reaction at 49 °C when [Fe²+] = 3.80 M and [Mg²+] = 0.210 M. Express your answer to three significant figures and include the appropriate units. ▸ View Available Hint(s) μA E= Value Submit y Units Mg(s) + Fe²+ (aq) →Mg²+ (aq) + Fe(s) ?
What is the cell potential for the reaction at 49 °C when [Fe²+] = 3.80 M and [Mg²+] = 0.210 M. Express your answer to three significant figures and include the appropriate units. ▸ View Available Hint(s) μA E= Value Submit y Units Mg(s) + Fe²+ (aq) →Mg²+ (aq) + Fe(s) ?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Learning Goal:
To learn how to use the Nernst equation.
The standard reduction potentials listed in any reference table are only valid at
the common reference temperature of 25 °C and standard conditions of 1 M
for solutions and 1 atm for gases. To calculate the cell potential at nonstandard
conditions, one uses the Nernst equation,
E=E 2.303 RT 1
nF
-log10 Q
where E is the potential in volts. Eo is the standard potential at 25°C in volts.
R 8.314 J/(K-mol) is the gas constant, T is the temperature in kelvins.
12 is the number of moles of electrons transferred, F = 96, 500 C/(mol e-)
is the Faraday constant, and is the reaction quotient.
At the common reference temperture of 298 K. substituting each constant into
the equation the result is
E = E-
Vlog 10 Q
0.0592 V
71](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f1d9a89-71fd-4cbf-9dfd-d06b2874d66e%2Fed4a2c79-2bed-4736-8820-3b447aae0b07%2Fwmw4mdg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Learning Goal:
To learn how to use the Nernst equation.
The standard reduction potentials listed in any reference table are only valid at
the common reference temperature of 25 °C and standard conditions of 1 M
for solutions and 1 atm for gases. To calculate the cell potential at nonstandard
conditions, one uses the Nernst equation,
E=E 2.303 RT 1
nF
-log10 Q
where E is the potential in volts. Eo is the standard potential at 25°C in volts.
R 8.314 J/(K-mol) is the gas constant, T is the temperature in kelvins.
12 is the number of moles of electrons transferred, F = 96, 500 C/(mol e-)
is the Faraday constant, and is the reaction quotient.
At the common reference temperture of 298 K. substituting each constant into
the equation the result is
E = E-
Vlog 10 Q
0.0592 V
71
![What is the cell potential for the reaction
at 49 °C when [Fe²+] = 3.80 M and [Mg²+] = 0.210 M.
Express your answer to three significant figures and include the appropriate units.
▸ View Available Hint(s)
μA
E= Value
Submit
y
Units
Mg(s) + Fe²+ (aq) →Mg²+ (aq) + Fe(s)
?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f1d9a89-71fd-4cbf-9dfd-d06b2874d66e%2Fed4a2c79-2bed-4736-8820-3b447aae0b07%2Fj14857k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:What is the cell potential for the reaction
at 49 °C when [Fe²+] = 3.80 M and [Mg²+] = 0.210 M.
Express your answer to three significant figures and include the appropriate units.
▸ View Available Hint(s)
μA
E= Value
Submit
y
Units
Mg(s) + Fe²+ (aq) →Mg²+ (aq) + Fe(s)
?
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