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- Calculate the equilibrium constant for the following reaction at 25 °C, 2 103(aq) +5 Hg() + 12 H¹ (aq) →→ 12(s) + 5 Hg2+ (aq) + 6 H₂O(l) The standard reduction potentials are as follows: ->> IOş(aq) +6 H*(aq)+5 e→ I,(s)+3H,O) Hg2+ (aq) +2e →→ Hg() a. 6 × 105 b. 3 x 1011 c. 6 × 1028 d. 3 × 1057 e. 3 × 10-58 emies? E = +1.20 V E = +0.86 V

- Calculate the equilibrium constant for the following reaction at 25 °C, 2 103(aq) +5 Hg() + 12 H¹ (aq) →→ 12(s) + 5 Hg2+ (aq) + 6 H₂O(l) The standard reduction potentials are as follows: ->> IOş(aq) +6 H*(aq)+5 e→ I,(s)+3H,O) Hg2+ (aq) +2e →→ Hg() a. 6 × 105 b. 3 x 1011 c. 6 × 1028 d. 3 × 1057 e. 3 × 10-58 emies? E = +1.20 V E = +0.86 V

Chemistry
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 the equilibrium constant for the following reaction at 25 °C, 2 103 (aq) + 5 Hg() + 12 H¹ (aq) →→ 12(s) + 5 Hg2+ (aq) + 6 H₂O(l) The standard reduction potentials are as follows: ->> IO3(aq) + 6 H(aq) + 5 e→ 1₂(s) + 3 H₂O(l) Hg2+ (aq) + 2e →→ Hg() a. 6 × 105 b. 3 × 1011 c. 6 × 1028 d. 3 × 1057 X e. 3 × 10-58 Juneming? E = +1.20 V E = +0.86 V 86
Transcribed Image Text:- Calculate the equilibrium constant for the following reaction at 25 °C, 2 103 (aq) + 5 Hg() + 12 H¹ (aq) →→ 12(s) + 5 Hg2+ (aq) + 6 H₂O(l) The standard reduction potentials are as follows: ->> IO3(aq) + 6 H(aq) + 5 e→ 1₂(s) + 3 H₂O(l) Hg2+ (aq) + 2e →→ Hg() a. 6 × 105 b. 3 × 1011 c. 6 × 1028 d. 3 × 1057 X e. 3 × 10-58 Juneming? E = +1.20 V E = +0.86 V 86
Write a balanced half-reaction for the reduction of CrO2 (aq) to Cr(OH)3(s) in a basic solution. a. CrO42 (aq) + 3 H¹ (aq) + 3 e → Cr(OH)3(s) b. CrO4 (aq) + 3 OH(aq) → Cr(OH)3(s) + 2 O₂(g) c. CrO4 (aq) + 3 H (aq) →→→ Cr(OH)3(s) + 2 e¯ X 3 d. CrO2 (aq) + 4H₂O() + 3 e→Cr(OH)3(s) + 5 OH(aq) e. CrO2 (aq) + 3 OH(aq) + 3 e→Cr(OH)3(s) + 2 O₂(g)
Transcribed Image Text:Write a balanced half-reaction for the reduction of CrO2 (aq) to Cr(OH)3(s) in a basic solution. a. CrO42 (aq) + 3 H¹ (aq) + 3 e → Cr(OH)3(s) b. CrO4 (aq) + 3 OH(aq) → Cr(OH)3(s) + 2 O₂(g) c. CrO4 (aq) + 3 H (aq) →→→ Cr(OH)3(s) + 2 e¯ X 3 d. CrO2 (aq) + 4H₂O() + 3 e→Cr(OH)3(s) + 5 OH(aq) e. CrO2 (aq) + 3 OH(aq) + 3 e→Cr(OH)3(s) + 2 O₂(g)
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