Open Question
The standard reduction potential of Eu2+(aq) is -0.43 V. Using Appendix E, which of the following substances is capable of reducing Eu3+(aq) to Eu2+(aq) under standard conditions: Al, Co, H2O2, N2H5+, H2C2O4?
Ch.20 - Electrochemistry
Chapter 20, Problem 51c
Given the following reduction half-reactions:
Fe3+(aq) + e- → Fe2+(aq) E°red = +0.77 V
S2O62-(aq) + 4 H+(aq) + 2 e- → 2 H2SO3(aq) E°red = +0.60 V
N2O(g) + 2 H+(aq) + 2 e- → N2(g) + H2O(l) E°red = -1.77 V
VO2+(aq) + 2 H+(aq) + e- → VO2+ + H2O(l) E°red = +1.00 V
(c) Calculate the equilibrium constant K for each reaction at 298 K.
Verified step by step guidance1
Identify the standard reduction potential (E°) for each half-reaction provided in the problem.
Use the Nernst equation in its simplified form for standard conditions: E°cell = (RT/nF) * ln(K), where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298 K), n is the number of moles of electrons transferred, and F is Faraday's constant (96485 C/mol).
Rearrange the Nernst equation to solve for the equilibrium constant K: ln(K) = nF * E°cell / RT.
For each half-reaction, determine the number of electrons transferred (n) and substitute the values of E°cell, R, T, and F into the rearranged equation to calculate ln(K).
Exponentiate the result from the previous step to find the equilibrium constant K for each reaction: K = e^(ln(K)).
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Textbook Question
Given the following reduction half-reactions:
Fe3+(aq) + e- → Fe2+(aq) E°red = +0.77 V
S2O62-(aq) + 4 H+(aq) + 2 e- → 2 H2SO3(aq) E°red = +0.60 V
N2O(g) + 2 H+(aq) + 2 e- → N2(g) + H2O(l) E°red = -1.77 V
VO2+(aq) + 2 H+(aq) + e- → VO2+ + H2O(l) E°red = +1.00 V
(a) Write balanced chemical equations for the oxidation of Fe2+(aq) by S2O62-(aq), by N2O(aq), and by VO2+(aq).
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Textbook Question
Given the following reduction half-reactions:
Fe3+(aq) + e- → Fe2+(aq) E°red = +0.77 V
S2O62-(aq) + 4 H+(aq) + 2 e- → 2 H2SO3(aq) E°red = +0.60 V
N2O(g) + 2 H+(aq) + 2 e- → N2(g) + H2O(l) E°red = -1.77 V
VO2+(aq) + 2 H+(aq) + e- → VO2+ + H2O(l) E°red = +1.00 V
(b) Calculate ∆G° for each reaction at 298 K.
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