andard poY³+ (aq) = X³+ (aq) + Y(s)HaJK = 4.04 x 10-5a 7 of 14Macmillan Learning>Calculate the equilibrium constant, K, for the reaction shown at 25 °C.Fe³+ (aq) + B(s) + 6H₂O(1)→ Fe(s) + H₂ BO3(s) + 3 H₂O+ (aq)The balanced reduction half-reactions for the equation and their respective standard reduction potential values (E") areFe³+ (aq) + 3e →→→ Fe(s)H₂BO3(s) + 3H₂O+ (aq) + 3e¯K =->>B(s) + 6H₂O(1)HaE° = -0.04 VE = -0.8698 V09/12/2

Given , Reaction : Fe3+(aq) + B(s) +6 H2O(I) →FeS(s) + H3BO3(s) + 3H3O+(aq)

Macmillan Learning Calculate the equilibrium constant, K, for the reaction shown at 25 °C. Fe³+ (aq) + B(s) + 6H₂O(1) — Fe(s) + H,BO,(s) + 3H,O*(aq) The balanced reduction half-reactions for the equation and their respective standard reduction potential values (E") are Fe³+ (aq) + 3e →→ Fe(s) H₂BO3(s) + 3H₂O+ (aq) + 3 e K = -> B(s) + 6H₂O(1) E = -0.04 V E-0.8698 V

Macmillan Learning Calculate the equilibrium constant, K, for the reaction shown at 25 °C. Fe³+ (aq) + B(s) + 6H₂O(1) — Fe(s) + H,BO,(s) + 3H,O*(aq) The balanced reduction half-reactions for the equation and their respective standard reduction potential values (E") are Fe³+ (aq) + 3e →→ Fe(s) H₂BO3(s) + 3H₂O+ (aq) + 3 e K = -> B(s) + 6H₂O(1) E = -0.04 V E-0.8698 V

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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