identify the reducing and the oxidizing agent for each reaction:
a) 2na + br2 = 2nabr
b) h2 + cl2 = 2hcl
c) 2li + f2 = 2lif
d) s + cl2 = scl2
e) n2 + 2o2 = 2no2
f) mg + cu(no3)2 = mg(no3)2 + cu
a) In the reaction 2Na + Br2 = 2NaBr, sodium (Na) is the reducing agent since it is oxidized from its elemental form, Na, to Na+ in NaBr. Bromine (Br2) is the oxidizing agent since it is reduced from Br2 to Br- in NaBr.
b) In the reaction H2 + Cl2 = 2HCl, hydrogen (H2) is the reducing agent since it is oxidized from H2 to H+ ions in HCl. Chlorine (Cl2) is the oxidizing agent since it is reduced from its elemental form, Cl2, to Cl- ions in HCl.
c) In the reaction 2Li + F2 = 2LiF, lithium (Li) is the reducing agent since it is oxidized from its elemental form, Li, to Li+ ions in LiF. Fluorine (F2) is the oxidizing agent since it is reduced from its elemental form, F2, to F- ions in LiF.
d) In the reaction S + Cl2 = SCl2, sulfur (S) is the reducing agent since it is oxidized from its elemental form, S, to S2+ ions in SCl2. Chlorine (Cl2) is the oxidizing agent since it is reduced from its elemental form, Cl2, to Cl- ions in SCl2.
e) In the reaction N2 + 2O2 = 2NO2, nitrogen (N2) is the reducing agent since it is oxidized from N2 to N+ ions in NO2. Oxygen (O2) is the oxidizing agent since it is reduced from its elemental form, O2, to O- ions in NO2.
f) In the reaction Mg + Cu(NO3)2 = Mg(NO3)2 + Cu, magnesium (Mg) is the reducing agent since it is oxidized from its elemental form, Mg, to Mg2+ ions in Mg(NO3)2. Copper (Cu) in Cu(NO3)2 is the oxidizing agent since it is reduced from Cu2+ ions to Cu atoms in Cu.
To determine the reducing and oxidizing agents in a redox reaction, you need to assign oxidation numbers to each element in the reactants and products. The element that undergoes oxidation is the reducing agent, and the element that undergoes reduction is the oxidizing agent.
Let's go through each reaction and determine the reducing and oxidizing agents:
a) 2Na + Br2 = 2NaBr
In this reaction, the oxidation number of Na changes from 0 to +1, and the oxidation number of Br changes from 0 to -1. The reducing agent is Na (it is oxidized from 0 to +1) and the oxidizing agent is Br2 (it is reduced from 0 to -1).
b) H2 + Cl2 = 2HCl
In this reaction, the oxidation number of H remains unchanged at 0, and the oxidation number of Cl changes from 0 to -1. The reducing agent is H2 (even though it doesn't undergo oxidation because its oxidation number remains unchanged) and the oxidizing agent is Cl2 (it is reduced from 0 to -1).
c) 2Li + F2 = 2LiF
In this reaction, the oxidation number of Li changes from 0 to +1, and the oxidation number of F changes from 0 to -1. The reducing agent is Li (it is oxidized from 0 to +1) and the oxidizing agent is F2 (it is reduced from 0 to -1).
d) S + Cl2 = SCl2
In this reaction, the oxidation number of S changes from 0 to +4, and the oxidation number of Cl changes from 0 to -1. The reducing agent is Cl2 (it is reduced from 0 to -1) and the oxidizing agent is S (it is oxidized from 0 to +4).
e) N2 + 2O2 = 2NO2
In this reaction, the oxidation number of N changes from 0 to +4, and the oxidation number of O changes from 0 to -2. The reducing agent is N2 (it is oxidized from 0 to +4) and the oxidizing agent is O2 (it is reduced from 0 to -2).
f) Mg + Cu(NO3)2 = Mg(NO3)2 + Cu
In this reaction, the oxidation number of Mg changes from 0 to +2, and the oxidation number of Cu changes from +2 to 0. The reducing agent is Mg (it is oxidized from 0 to +2) and the oxidizing agent is Cu(NO3)2 (it is reduced from +2 to 0).
Remember, the key is to identify the change in oxidation numbers. The element that is oxidized is the reducing agent, and the element that is reduced is the oxidizing agent.
OIL RIG= Oxidation is losing and reduction is gaining. The one being oxidized is the reducing agent and the one being reduced is the oxidizing agent.
Maybe Dr. Bob222 will do this one for you.