1. 2IO3- + 10Ag+ 12H+I2 + 10Ag++ 6H2O
In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.
2. Write a net ionic equation for the reaction that occurs when aqueous solutions of nitric acid and ammonia are combined.
3. The following compounds are water soluble. What ions are produced when the compounds are in aqueous solutions? Enter each ion preceded by the number produced upon dissolution.
FeCl2: cation and anion?
MgSO4: cation and anion?
1.
To identify the element oxidized, we need to compare the oxidation state of the element before and after the reaction. In this reaction, the oxidation state of iodine (I) changes from -1 in 2IO3- to 0 in I2, so iodine is being reduced.
To identify the element reduced, we follow the same process. In this reaction, the oxidation state of silver (Ag) changes from +1 in 10Ag+ to +2 in 10Ag++, so silver is being oxidized.
The oxidizing agent is the species that causes another element to be oxidized. In this reaction, Ag+ is causing I to be oxidized, so Ag+ is the oxidizing agent.
The reducing agent is the species that causes another element to be reduced. In this reaction, I- is causing Ag+ to be reduced, so I- is the reducing agent.
2.
The net ionic equation for the reaction between nitric acid (HNO3) and ammonia (NH3) can be written as:
H+ (aq) + NH3 (aq) → NH4+ (aq)
3.
FeCl2 dissociates into Fe2+ and 2 Cl- ions in aqueous solution.
MgSO4 dissociates into Mg2+ and SO4 2- ions in aqueous solution.
1. To identify the element that is oxidized and reduced in the given redox reaction, we need to determine the changes in oxidation numbers for each element.
First, let's assign oxidation numbers for each element involved in the reaction:
- Oxygen typically has an oxidation number of -2.
- Hydrogen usually has an oxidation number of +1.
- Silver (Ag) typically has an oxidation number of +1.
- Iodine (I) can have various oxidation numbers depending on the compound. In this case, it has an oxidation number of 0 in I2.
Using these guidelines, we can assign oxidation numbers to the elements in the reactants and products:
2IO3-: Let's assign the oxidation number of iodine (I) in IO3- as x. Since there are three oxygen atoms each with an oxidation number of -2, we can write the equation as 3(-2) + x = -1 (the overall charge of IO3-). Solving this equation gives us x=+5. Thus, the oxidation number of iodine in IO3- is +5.
10Ag+: Ag has an oxidation number of +1, so the total oxidation number of Ag+ is +10.
12H+: Since hydrogen usually has an oxidation number of +1, the total oxidation number of H+ is +12.
On the product side:
I2: Iodine (I) has an oxidation number of 0.
10Ag++: Each Ag+ has an oxidation number of +1, so the total oxidation number of Ag++ is +10.
6H2O: Each oxygen atom has an oxidation number of -2, so the total oxidation number of oxygen in H2O is 6(-2)=-12. Since hydrogen typically has an oxidation number of +1, the total oxidation number of hydrogen in H2O is 6(+1)=+6.
From the changes in oxidation numbers, we can identify the element oxidized, reduced, and the oxidizing/reducing agents:
- The element oxidized is Iodine (I), which changes from an oxidation number of +5 to 0, indicating a reduction in oxidation number.
- The element reduced is Silver (Ag), which changes from an oxidation number of +1 to 0, indicating an increase in oxidation number.
- The oxidizing agent is IO3-, as it causes iodine to be reduced.
- The reducing agent is Ag+, as it causes silver to be oxidized.
2. To write the net ionic equation for the reaction between nitric acid (HNO3) and ammonia (NH3), we first need to write the balanced molecular equation. The reaction can be represented as:
HNO3 + NH3 → NH4NO3
In this reaction, nitric acid (HNO3) reacts with ammonia (NH3) to produce ammonium nitrate (NH4NO3).
Next, we need to identify the strong electrolytes and write the dissociation equations for them. Nitric acid and ammonium nitrate completely dissociate in aqueous solution, while ammonia is a weak base and remains mostly undissociated. The dissociation equations are as follows:
HNO3 → H+ + NO3-
NH4NO3 → NH4+ + NO3-
Since both nitric acid and ammonium nitrate produce the same ions (NO3-) in aqueous solution, they are called spectator ions and do not participate in the net ionic equation.
The net ionic equation is obtained by removing the spectator ions from the balanced equation:
H+ + NH3 → NH4+
Note that in the net ionic equation, the nitrate ions (NO3-) from both reactants have canceled out.
3. When FeCl2 (Iron(II) chloride) is dissolved in water, it dissociates into two ions:
Fe2+ and 2Cl-
- The cation produced upon dissolution is Fe2+ (Iron(II) ion).
- The anion produced upon dissolution is Cl- (Chloride ion).
Similarly, when MgSO4 (Magnesium sulfate) is dissolved in water, it dissociates into three ions:
Mg2+ and 1SO42-
- The cation produced upon dissolution is Mg2+ (Magnesium ion).
- The anion produced upon dissolution is SO42- (Sulfate ion).
1. You MUST put in an arrow.
I changes from +5 on the left to 0 on the right.
Ag changes from +1 on the left to +3 on the right for each Ag; howewver, you must have made up a problem because Ag^2+ is not common.
Count up electrons lost and gained for the above, then
Loss of e is oxidation; gain of e is reduction.
The element oxidized is the reducing agent.
The element reduced is the oxidizing agent.
2. HNO3(aq) + NH3(aq) ==> NH4NO3(aq)
net ionic is H^+ + NH3 ==> NH4^+
3. FeCl2==> Fe^2++ 2Cl^-
MgSO4 ==> Mg^2+ + [SO4]^2-