The Nernst EquationA nonstandard cell or half-cell potential can be calculated using the Nernst equation:RTE = E°-InQ%3DnFwhereE = potential under nonstandard conditions%3DE = standard potentialR = ideal gas constant = 8.314 J/K mol%3DT = kelvin temperature%3Dn = number of moles of electrons for the reaction aswrittenF = charge carried by 1 mol of electrons =96, 485 C/mol = 96, 485 J/V. mol%3D%3DQ = reaction quotient%3DIt is customary to use the equation in a form where numberical values are substituted for R, T, and F at a temperature of 25° C.(8.314 J/K mol)(298K)96, 485 J/V. molRT= 0.02568 Vand the Nernst equation with the potentials in volts is:0.02568In QE = E° -nSometimes it is more convenient to use base-10 logarithms, and the substitution of 2.3026 log for In is made.(2.3026 x 0.02568 = 0.0591)%3DE = E0.0591logQA common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correctlogarithm.Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction:Ni²+ (aq) + Cu(s) → Ni(s) + Cu²+ (aq)Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.K =%3DAG° for this reaction would bethan zero.greaterlessRetry Entire GreSubmit Answerbre aroun attemnts remainina

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A common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm. Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction: Ni2+ (aq) + Cu(s) Ni(s) + Cu+ (aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. K = AG° for this reaction would be v than zero. greater less Retry Entire Grd Submit Answer bre arnun attemnte remaininn

A common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm. Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction: Ni2+ (aq) + Cu(s) Ni(s) + Cu+ (aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. K = AG° for this reaction would be v than zero. greater less Retry Entire Grd Submit Answer bre arnun attemnte remaininn

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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Calculate K (both parts of the question)
Calculate the cell potential, E, for the given reactions at 25.00 "C using the ion concentrations provided. Then, determine if th cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E). Pt(s) + Fe²(aq) = P² (aq) + Fe(s) [Fe²*1 = 0.0064 M E= incorrect -4761 E= Cu(s) + 2 Ag* (aq) = Cu² (aq) + 2 Ag(s) |Cu²+1 = 0.027 M Ag1=0.027 M E [P1²*1 = 0.067 M Co² (aq) + Ti (aq) — Co¹ (aq) + Ti² '(aq) Co²1 = 0.055 M Co= 0.030 M ITP 1 = 0.0070 M T² = 0.0107 M mam V The cell is not spontaneous spontaneous. The cell is spontaneous not spontaneous. incorrect The cell is spontaneous not spontaneoux.
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Be sure your answer has the correct number of significant digits. 2+ 3+ Mn²+ (aq) + 2H₂O (1) +2Fe³+ (aq) → MnO₂ (s) + 4H* (aq) +2Fe²+ (aq) kJ x10 [:] OL A பா
Use standard reduction potentials to calculate the equilibrium constant for the reaction: Hg2+(aq) + 2Cu+(aq)Hg(1) + 2Cu2+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: AG° for this reaction would be than zero.
K124.
Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kJ for the reaction: Cd2+ (aq) + Zn(s)→→→→→→ Cd(s) + Zn²+ (aq) Answer: kJ K for this reaction would be than one.
What is the equilibrium constant for the following reaction at 25 °C? Round your answer to 1 decimal place. 2 Cu(s) + 2 Ag+ (aq) ⇒ Cu²+ (aq) + 2 Ag (s) = Note: Reference the Standard reduction potentials at 25 °C table for additional information. K = = x10
Use standard reduction potentials to calculate the equilibrium constant for the reaction: Cu²+ (aq) + Pb(s) → Cu(s) + Pb²+ (aq) 2+ Cu²+ (aq) + 2e Pb²+ (aq) + 2e → Cu(s) Fre 0.337 V → Pb(s) Fe -0.126 V red = Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant at 298 K: AGO for this reaction would be | than zero.
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy following redox reaction. Be sure your answer has the correct number of significant digits. H₂(g) + 2OH¯(aq) + 2Fe³+ (aq) → 2H₂O (1) + 2Fe²+ (aq) kJ x10 X Ś AGO for the
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Cu²+ (aq) + 2Cu+ (aq) → Cu(s) + 2Cu²+ (aq) 2+ Cu²+ (aq) + 2e → Cu(s) Ee = 0.337 V red 0.153 V = 2+ Cu²+ (aq) +e → Cut (aq) E red AG° kJ K for this reaction would be than one.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2H+ (aq) + Hg(1)→ H₂(g) + Hg²+ (aq) 2H+ (aq) + 2e¯ → H₂(g) Fº red 2+ Hg2 (aq) + 2e¯ → Hg(1) AGO = || kJ K for this reaction would be Eº red = 0.000 V = = 0.855 V than one.
Use standard reduction potentials to calculate the equilibrium constant for the reaction: (aq) + Ag(s) → Fe²+ (s) + Ag+ (aq) Fe³+ From the table of standard reduction potentials: = 0.771 V, E E3+ = 0.799 V Ag+ /Ag Fe³+/Fe²+ Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant at 298 K: AGO for this reaction would be Submit Answer than zero.