What is the cell potential for the reaction at 89 °C when [Fe2+] = 2.80 M and [Mg2+] = 0.110 M. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) E= 2.86 V MO ? Mg(s) + Fe²+ (aq)-Mg2+ (aq) + Fe(s) Submit Previous Answers Request Answer

What is the cell potential for the reaction at 89 °C when [Fe2+] = 2.80 M and [Mg2+] = 0.110 M. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) E= 2.86 V MO ? Mg(s) + Fe²+ (aq)-Mg2+ (aq) + Fe(s) Submit Previous Answers Request Answer

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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Question

What is the cell potential for the reaction
at 89 °C when [Fe²+] = 2.80 M and [Mg2+] = 0.110 M.
Express your answer to three significant figures and include the appropriate units.
▸ View Available Hint(s)
E= 2.86
Submit
V
Previous Answers Request Answer
?
2+
Mg(s) + Fe²+ (aq) →→Mg²+ (aq) + Fe(s)
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When the Agt concentration is 1.01 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.072V. What is the Mn2+ concentration? 2Ag+(aq) + Mn(s)→2A9(s) + Mn2+(aq) Answer: M
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What is the answer?
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
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+ Cell Potential and Equilibrium The equilibrium constant, K, for a redox reaction is related to the standard potential, E, by the equation In K FE RT where 72 is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e-). R (the gas constant) is equal to 8.314 J/(mol K), and T is the Kelvin temperature. Standard reduction potentials ▾ Part A Use the table Express your answer numerically. ▸ View Available Hint(s) K= ΥΠΙ ΑΣΦ 9 Reduction half-reaction Ag+ (aq) +eAg(s) Cu²+ (aq) +201 >Cu(s) Sn+ (aq) + 4e →→Sn(s) 2H+ (aq) +2e →H₂(g) Ni2+ (aq) + 2e →Ni(s) Fe²+ (aq) + 2e →Fe(s) Zn²+ (aq) +2e →Zn(s) Al³+ (aq) + 3e →Al(s) Mg²+ (aq) + 2e →Mg(s) standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 °C) for the following reaction: Fe(s) + Ni²+ (aq) →>Fe²+ (aq) + Ni(s) ? E (V) 0.80 0.34 0.15 0 0.26 -0.45 0.76 -1.66 2.37 29 o Review | Constants | Period
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