Treatment of gold metal with BrF3 and KF produces ...

Question

Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold.Identify the oxidizing agent in this reaction.

Jules Bruno · Accepted Answer

We’re being asked to identify the oxidizing agent in the given reaction:Au + BrF3 + KF → Br2 + KAuF4To do so, we need to identify the change in oxidation state for each element in the reaction. Recall the mnemonic for redox reactions:Loss Electrons → Oxidation → Reducing AgentGain Electrons → Reduction → Oxidizing AgentThis means the element that gains electrons/reduced is the oxidizing agent.[readmore]The rules for oxidation states are as follows:A. General Rules:1. For an atom in its elemental form,  O.S. = 0     2. For an ion (Li+, Al3+, etc.) O.S. = chargeB. Specific Rules:1. Group 1A              O.S. = +12. Group 2A              O.S. = +23. Hydrogen              O.S. = +1, with nonmetals                                  O.S. = –1 with metals and boron4. Fluorine                 O.S. = –15. Oxygen                  O.S. = –1 in peroxides (X2O2, X = Group 1A)                                  O.S. = –1/2 in superoxides (XO2, X = Group 1A)                                  O.S. = –2 in other compounds6. Group 7A              O.S. = –1 (except when bonded to O)Reactants:For AuThis is an atom in its elemental form, which means  O.S. of Au = 0.For BrF3: Fluorine (F) belongs to Group 7A which means  O.S. of F = –1.This is also a neutral compound so the total oxidation state is zero.O.S. of Br + O.S. of F = 0O.S. of Br + [3 x (–1)] = 0O.S. of Br= +3For KF: Fluorine (F) belongs to Group 7A which means  O.S. of F = –1.Potassium (K) belongs to Group 1A which means  O.S. of K = +1.Products:For Br2: This is an atom bonded into itself, which means O.S. of Br= 0.For KAuF4: Fluorine (F) belongs to Group 7A which means  O.S. of F = –1.Potassium (K) belongs to Group 1A which means  O.S. of K = +1.This is also a neutral compound so the total oxidation state is zero.O.S. of K + O.S. of Au + O.S. of F = 0[+1] + O.S. of Au + [4 x (–1)] = 0+1 + O.S. Of Au -4 = 0-1                      +4     -1 + 4O.S. of Au = +3Changes in O.S.:Br:      +3 → 0           decreased/gained electronsF:        –1 → –1         no changeAu:        0 → +3           increased/lost electronsK:        +1 → +1         no changeGold (Au) lost electrons, which means it was oxidized and is the reducing agent. On the other hand, Br gained electrons, meaning it was reduced and is the oxidizing agent.

Problem: Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold.Identify the oxidizing agent in this reaction.

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Problem Details

Problem: Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold.Identify the oxidizing agent in this reaction.

FREE Expert Solution

Problem Details

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