Question

note that for each of the four reactions, one of the gray boxes will be left blank and the other will be filled with electron(s). use the symbol e⁻ to represent an electron.
+fe²⁺(aq)→fe³⁺(aq)+
choose one:
oxidation
reduction
part 2 (2 points)
+agi(s)→ag(s)+i⁻(aq)+
choose one:
oxidation
reduction
part 3 (2 points)
+vo₂⁺(aq)+2h⁺(aq)→vo²⁺(aq)+h₂o(l)+

Explanation:

Step1: Analyze oxidation - state change in Part 1

In the reaction $\text{Fe}^{2 +}(aq)\to\text{Fe}^{3+}(aq)$, the iron ion loses an electron. The oxidation - state of iron increases from +2 to +3. Oxidation is the loss of electrons. So the reaction is an oxidation reaction. The half - reaction is $\text{Fe}^{2+}(aq)\to\text{Fe}^{3+}(aq)+e^-$.

Step2: Analyze oxidation - state change in Part 2

In the reaction $\text{AgI}(s)\to\text{Ag}(s)+\text{I}^-(aq)$, silver in $\text{AgI}$ has an oxidation state of + 1, and in $\text{Ag}(s)$ it has an oxidation state of 0. Silver gains an electron, and reduction is the gain of electrons. The half - reaction is $\text{AgI}(s)+e^-\to\text{Ag}(s)+\text{I}^-(aq)$.

Step3: Analyze oxidation - state change in Part 3

In the reaction $\text{VO}_2^+(aq)+2\text{H}^+(aq)\to\text{VO}^{2 +}(aq)+\text{H}_2\text{O}(l)$, the oxidation state of vanadium in $\text{VO}_2^+$ is +5 and in $\text{VO}^{2+}$ is +4. Vanadium gains an electron. So it is a reduction reaction. The half - reaction is $\text{VO}_2^+(aq)+2\text{H}^+(aq)+e^-\to\text{VO}^{2+}(aq)+\text{H}_2\text{O}(l)$.

Answer:

Part 1: Oxidation, $\text{Fe}^{2+}(aq)\to\text{Fe}^{3+}(aq)+e^-$
Part 2: Reduction, $\text{AgI}(s)+e^-\to\text{Ag}(s)+\text{I}^-(aq)$
Part 3: Reduction, $\text{VO}_2^+(aq)+2\text{H}^+(aq)+e^-\to\text{VO}^{2+}(aq)+\text{H}_2\text{O}(l)$