Use the References to access important values if needed for this question. Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Ag (aq) +21 (aq) → 2Ag(s) + 1₂ (8) Ag+ (aq) +e → Ag(s) E red 12 (8)+2e21 (ag) E red Submit Answer = 0.799 V Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant at 298 K: AG for this reaction would be Retry Entire Group = 0.535 V e than zero. 9 more group attempts remaining

Use the References to access important values if needed for this question. Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Ag (aq) +21 (aq) → 2Ag(s) + 1₂ (8) Ag+ (aq) +e → Ag(s) E red 12 (8)+2e21 (ag) E red Submit Answer = 0.799 V Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant at 298 K: AG for this reaction would be Retry Entire Group = 0.535 V e than zero. 9 more group attempts remaining

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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Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Ag+ (aq) + 21¯(aq) → 2Ag(s) + I₂ (s) Ag+ (aq) + e¯ → Ag(s) E = 0.799 V red I₂ (s) + 2e¯ → 21¯ (aq) Eº 0.535 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.
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Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Pb²+ (aq) + Cu(s) - → Pb(s) + Cu²+ (aq) AG° Pb²+ (aq) + 2e¯¯ → Pb(s) Ee Cu²+ (aq) + 2e¯ → Cu(s) Fre red = kJ K for this reaction would be = -0.126 V = 0.337 V + than one.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2Cu²+ u²+ (aq) + 21- (aq) → 2Cut (aq) + 12 (8) 2+ Cu²+ (aq) + e → Cut (aq) Ee red I2 (s) +2e → 21¯ (aq) AGO = K for this reaction would be kJ E red - 0.153 V 0.535 V than one.
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
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Round your answer to 3 significant digits. 4Br (aq) +N2 (g)+4H₂O (1) → 2Bг2 (1)+N₂H4 (aq) +40H (aq) OkJ T x10 ×
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Ag (aq) + Fe(s) → 2Ag(s) + Fe²+ (aq) Ag+ (aq) + e → Ag(s) E red Fe2+ (aq) + 2e → Fe(s) E red www 0.799 V -0.440 V 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.
Standard Electrode Potentials at 25 °C Reduction Half-Reaction 2 H₂O(0+2 e Mn²+ (aq) + 2 e Al³+ (aq) + 3 e. Mg²+ (aq) +2 e Na (aq) + e Ca²+ (aq) + 2 e Bat (aq) +2 e K¹(aq) + e consider the following redox reaction: 2 Al(s) +3 Mg2+ (aq)-2 Al³+ (aq) + 3 Mg(s) 1) What is the Eº for the reaction? 2) Calculate AG for the redox reaction at 25°C. The answer to 1) is [Select] The answer to 2) is [Select] A →→H₂(g) + 2 OH(aq) - Mn(s) -Al(s) Mg(s) -Na(s) -Ca(s) -Ba(s) -K(s) E° (V) -0.83 -1.18 -1.66 -2.37 -2.71 -2.76 -2.90 -2.92
Using data in the Table, what is the equilibrium constant for the reaction: 2Fe 3+(aq)+H 2(g) 2Fe 2+(aq) + 2H +(aq)?
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.
Enter electrons as e". Use smallest possible integer coefficients. States are not required. If a box is not needed, leave it blank. Standard Reduction Potentials (Volts) at 25 °C Cl₂ (g) + 2 e →2 Cl(aq) O2(g) + 4 H3O+ (aq) + 4 e 6 H₂O(1) 1.360 1.229 2 H2O(l) + 2e Al3+ (aq) + 3 e → Al (s) H2(g) + 2 OH(aq) -0.828 -1.660 A 0.114 M neutral aqueous AlCl3 solution is electrolyzed under 1 atm pressure using platinum electrodes. (a) Write the half-reactions expected to occur, taking into account the effect of the non-standard conditions on the electrode potentials. Half-reaction at anode: + + + + Half-reaction at cathode: + + + + (b) What is the expected decomposition potential?
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.