The dissolution reactions of gold (Au) in 1M HNO3 and 1M aqua regia (1M HNO3 + 1M HCI) are given below. Аu) + NOз (аq) + 4 H* » Au3+ + NO (g) (aq) + 2 H20(1) (1) (ад) Au(s) + NO3 + 4 H* (ag) + 4 Cl¯ (aq) (aq) → AuCl4¯ + NO(9) + 2 H20) (g) (2) (aq) Reduction Half – Reaction E° (V) Au3+ + 3 e- = Au 1.36 AuCl4 + 3 e¯ → Au + 4 Cl- 1.00 NO3- + 4 H* + 3 e¯ → NO + 2 H20 0.96 2 H30+ + 2 e 2 H2 + 2 H20 Calculate E°cel| and Keg for the dissolution of gold in 1M HNO3 under standard conditions. Calculate E°cell and Keg for the dissolution of gold in aqua regia under standard conditions. Using the values calculated in a) and b), determine the formation constant (K;) of AuCl4-. The formation reaction is given below. [Hint, express K; as a function of the Keq values from a) and b)]. Au3+ (аq) + 4 Cl (aq) → AuCl4 (aq) Important Information Atomic Masses (g/mol): H = 1.01, C = 12.01, N = 14.01, O = 16.00, %D Ne = 20.18, CI = 35.45, K = 39.10, Ar = 39.95, Cu = 63.55, Ag = 107.87 %3D %3D wwm Constants: R = 8.314 J/mol·K = 0.08206 L·atm/mol·K %3D ww m www m F = 96,485 C/mol = 96,485 J/V•mol %3D
The dissolution reactions of gold (Au) in 1M HNO3 and 1M aqua regia (1M HNO3 + 1M HCI) are given below. Аu) + NOз (аq) + 4 H* » Au3+ + NO (g) (aq) + 2 H20(1) (1) (ад) Au(s) + NO3 + 4 H* (ag) + 4 Cl¯ (aq) (aq) → AuCl4¯ + NO(9) + 2 H20) (g) (2) (aq) Reduction Half – Reaction E° (V) Au3+ + 3 e- = Au 1.36 AuCl4 + 3 e¯ → Au + 4 Cl- 1.00 NO3- + 4 H* + 3 e¯ → NO + 2 H20 0.96 2 H30+ + 2 e 2 H2 + 2 H20 Calculate E°cel| and Keg for the dissolution of gold in 1M HNO3 under standard conditions. Calculate E°cell and Keg for the dissolution of gold in aqua regia under standard conditions. Using the values calculated in a) and b), determine the formation constant (K;) of AuCl4-. The formation reaction is given below. [Hint, express K; as a function of the Keq values from a) and b)]. Au3+ (аq) + 4 Cl (aq) → AuCl4 (aq) Important Information Atomic Masses (g/mol): H = 1.01, C = 12.01, N = 14.01, O = 16.00, %D Ne = 20.18, CI = 35.45, K = 39.10, Ar = 39.95, Cu = 63.55, Ag = 107.87 %3D %3D wwm Constants: R = 8.314 J/mol·K = 0.08206 L·atm/mol·K %3D ww m www m F = 96,485 C/mol = 96,485 J/V•mol %3D
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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![The dissolution reactions of gold (Au) in 1M HNO3 and 1M aqua regia (1M HNO3 + 1M HCI) are
given below.
Аu) + NOз (аq)
+ 4 H*
» Au3+
+ NO (g)
(aq)
+ 2 H20(1)
(1)
(ад)
Au(s) + NO3
+ 4 H* (ag) + 4 Cl¯ (aq)
(aq)
→ AuCl4¯
+ NO(9) + 2 H20)
(g)
(2)
(aq)
Reduction Half – Reaction
E° (V)
Au3+ + 3 e- = Au
1.36
AuCl4 + 3 e¯ → Au + 4 Cl-
1.00
NO3- + 4 H* + 3 e¯ → NO + 2 H20
0.96
2 H30+ + 2 e 2 H2 + 2 H20
Calculate E°cel| and Keg for the dissolution of gold in 1M HNO3 under standard conditions.
Calculate E°cell and Keg for the dissolution of gold in aqua regia under standard conditions.
Using the values calculated in a) and b), determine the formation constant (K;) of AuCl4-.
The formation reaction is given below. [Hint, express K; as a function of the Keq values from a)
and b)].
Au3+
(аq)
+ 4 Cl (aq)
→ AuCl4
(aq)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe03749ab-2290-4f71-85cb-5c486db624e1%2F241c6acf-77d8-4382-9b37-40b9366683d1%2Fgg9aod_processed.png&w=3840&q=75)
Transcribed Image Text:The dissolution reactions of gold (Au) in 1M HNO3 and 1M aqua regia (1M HNO3 + 1M HCI) are
given below.
Аu) + NOз (аq)
+ 4 H*
» Au3+
+ NO (g)
(aq)
+ 2 H20(1)
(1)
(ад)
Au(s) + NO3
+ 4 H* (ag) + 4 Cl¯ (aq)
(aq)
→ AuCl4¯
+ NO(9) + 2 H20)
(g)
(2)
(aq)
Reduction Half – Reaction
E° (V)
Au3+ + 3 e- = Au
1.36
AuCl4 + 3 e¯ → Au + 4 Cl-
1.00
NO3- + 4 H* + 3 e¯ → NO + 2 H20
0.96
2 H30+ + 2 e 2 H2 + 2 H20
Calculate E°cel| and Keg for the dissolution of gold in 1M HNO3 under standard conditions.
Calculate E°cell and Keg for the dissolution of gold in aqua regia under standard conditions.
Using the values calculated in a) and b), determine the formation constant (K;) of AuCl4-.
The formation reaction is given below. [Hint, express K; as a function of the Keq values from a)
and b)].
Au3+
(аq)
+ 4 Cl (aq)
→ AuCl4
(aq)
![Important Information
Atomic Masses (g/mol): H = 1.01, C = 12.01, N = 14.01, O = 16.00,
%D
Ne = 20.18, CI = 35.45, K = 39.10, Ar = 39.95,
Cu = 63.55, Ag = 107.87
%3D
%3D
wwm
Constants:
R = 8.314 J/mol·K = 0.08206 L·atm/mol·K
%3D
ww m
www m
F = 96,485 C/mol = 96,485 J/V•mol
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe03749ab-2290-4f71-85cb-5c486db624e1%2F241c6acf-77d8-4382-9b37-40b9366683d1%2F2hmjcm_processed.png&w=3840&q=75)
Transcribed Image Text:Important Information
Atomic Masses (g/mol): H = 1.01, C = 12.01, N = 14.01, O = 16.00,
%D
Ne = 20.18, CI = 35.45, K = 39.10, Ar = 39.95,
Cu = 63.55, Ag = 107.87
%3D
%3D
wwm
Constants:
R = 8.314 J/mol·K = 0.08206 L·atm/mol·K
%3D
ww m
www m
F = 96,485 C/mol = 96,485 J/V•mol
%3D
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