1. Remember all of our questions pertain to the reaction below:
Br-(aq) + NO3- ==> NO(g) + Br2(l)
Options: Br, N, O, -2, -1, 0, +1, +2, -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5
[Select] is oxidized, it's oxidation number changes from [Select] to [Select].
[Select] is reduced, it's oxidation number changes from [Select] to [Select].
[Select]'s oxidation state doesn't change, its oxidation number is always [Select]
2. Fill the coefficients for the four original reactants and products:
Options: 8, 7, 6, 5, 4, 3, 2, 1, 0
[Select] Br- (aq) +
[Select] NO3- (aq) ==>
[Select] NO(g) +
[Select] Br2 (l)
3. If you were balancing this reaction in acidic conditions how many of each additional species would there be and where in the reaction would they be located.
Options: 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 and options: the reactant, products, neither
Protons (H+): There would be
[Select] on the [select] side
Water (H2O): There would be
[Select} on the [select] side
Hydroxide (OH-): There would be
[Select #] on the [select] side
4. The same thing as 3 but this time if you were balancing this reaction in basic conditions.
With no spacing this is really difficult to understand. I'm not all that interested in going through this in detail and answering everything. Let me get you started, then if you have trouble tell me the trouble you are having and what you don't understand and I can help you through it. In other words, I'm willing to HELP but I'm not willing to do all of the work for you. You need to learn how to do this.
Br-(aq) + NO3- ==> NO(g) + Br2(l)
Definitions you need to know. Oxidation is the loss of electrons; reduction is the gain of electrons. Br^- has an oxidation number (ON) of -1 on the left and zero on the right. So it lost e which makes it oxidized. N in NO3^- on the left has an ON of +5 and it changes to +2 on the right. That's a gain of electrons so it is reduced. The ON of Oxygen goesn't change. It is -2 on the left (for EACH O atom) and -2 on the right in NO.
The balanced equation is
6Br-(aq) + 2NO3 + 8H^+ ==> 2NO(g) + 3Br2(l) + 4H2O
I'll leave you here. I shall be happy to help you understand anything I've done above but you must be clear in what you don't understand. The same applies for the remaining questions.
1. To determine which species is oxidized and which is reduced, we need to assign oxidation numbers to each element in the reaction.
The oxidation state of an element is the hypothetical charge that an atom would have if all its bonds were 100% ionic.
In this reaction:
- Br- has an oxidation number of -1 since it is a halide ion.
- NO3- has an oxidation number of -1 as well, since it is a polyatomic ion and the sum of its oxidation numbers is -1.
- NO has an oxidation number of +2, since oxygen is usually assigned an oxidation number of -2, and the sum of the oxidation numbers for NO must be 0.
- Br2 has an oxidation number of 0, since it is a diatomic element and its oxidation number is always 0.
Based on these oxidation numbers, we can determine the answers:
- Br- is oxidized, its oxidation number changes from -1 to 0.
- NO3- is reduced, its oxidation number changes from -1 to +2.
- Oxygen's oxidation state doesn't change, its oxidation number is always -2.
2. To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.
The coefficients can be filled in as follows:
- 1 Br- (aq) +
- 2 NO3- (aq) ==>
- 2 NO (g) +
- 1 Br2 (l)
3. In acidic conditions, we need to add protons (H+) and water (H2O) to balance the equation. The protons usually come from an acid such as HCl, and the water is added to balance the oxygen atoms.
- There would be 10 protons (H+) on the reactant side.
- There would be 9 water molecules (H2O) on the product side.
- There would be no hydroxide ions (OH-) in acidic conditions.
The protons (H+) would be on the reactant side, and the water (H2O) would be on the product side.
4. In basic conditions, we need to add hydroxide ions (OH-) to balance the equation. The OH- ions react with the protons (H+) to form water (H2O).
- There would be 10 hydroxide ions (OH-) on the product side.
- There would be no protons (H+) or water (H2O) in basic conditions.
The hydroxide ions (OH-) would be on the product side.