Redox Reactions

What is the oxidation state of the metal cation in these compounds?

AgNO₃, ZnCl₂, Co₂O₃

1. Let's start with AgNO₃. NO₃ is an ion with a charge of -1, so Ag has an oxidation state of +1.

2. For ZnCl₂, we know that Cl usually has an oxidation state of -1. This is the case here, since there is no O or F in this compound. Since we have 2 mols of Cl to balance out Zn, then the ox. state of Zn is +2.

3. Lastly, for Co₂O₃, O₃ is an ion with a charge of -2. Since we need 2 mols of Co to balance, then the oxidation state of Co is +1.

In this reaction, which species is being oxidized, and which is being reduced?

$$Zn + CuSO_4 \rightarrow Cu + ZnSO_4$$

1. Let's start by identifying the oxidation states on the left side. Zn is a lone neutral element, so its oxidation state is 0.

2. SO₄ is an ion with a charge of -2, so Cu must have a charge of +2 since this is a neutral compound.

3. Now for the right side. Cu is now alone, so it's oxidation state is 0.

4. Zn is now bonded to SO₄, so it's oxidation state is +2.

So, our changes in oxidation state are Cu (+2 → 0) and Zn (0 → +2). Since copper's oxidation state has decreased, it is being reduced. That means that zinc is being oxidized since its oxidation state has increased.

Balance the redox reaction using the skeleton equation below:

$$Fe^{3+} + Sn^{2+} \rightarrow Fe^{2+} + Sn^{4+}$$

We already are given the reaction in its ionic form, so we start by determining the oxidation state for each species.

This step is pretty simple since all of our elements are ions, their oxidation numbers are just their charges.

Now we need to split this reaction into its half-reactions. Let's start with Fe.

$$Fe^{3+} \rightarrow Fe^{2+}$$

This half-reaction is the reduced half-reaction, since Fe's oxidation number is being reduced. Species being reduced gain an electron, so we want to balance our charge by adding an electron to the left side

$$Fe^{3+} + e^- \rightarrow Fe^{2+}$$

Now for our oxidation half-reaction.

$$Sn^{2+} \rightarrow Sn^{4+}$$

Species being oxidized lose an electron(s), so we balance by adding 2 electrons to the right side.

$$Sn^{2+} \rightarrow Sn^{4+} + 2e^-$$

Before we combine these reactions, we need to make sure that the electrons will cancel. We can multiply the reduction reaction by 2 so that our electrons will cancel.

$$2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$$

Now we can combine our reactions to get the final, balanced reaction

$$2Fe^{3+} + Sn^{2+} \rightarrow 2Fe^{2+} + Sn^{4+}$$

Write the balanced redox reaction (in acidic solution) using the skeleton formula below:

$$MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}$$

We start by determining each species' oxidation state:

Fe is simple since its oxidation state is just its charge, so it's oxidation state goes from +2 to +3. This means that Fe is being oxidized. The oxidation state of oxygen is -2, and we have 4 mols of it. This means that Mn must have an ox. state of +7 since the total charge of the compound is -1. Mn is being reduced since its oxidation state goes from +7 to +2.

Now we need to draw our half-reactions. Let's start with Mn

$$MnO_4^- \rightarrow Mn^{2+}$$

First, we must balance our oxygen by adding 4 mols of H₂O to the right side

$$MnO_4^- \rightarrow Mn^{2+} + 4H_2O$$

Next, we balance our hydrogen by adding 8 mols of H⁺ to the left side

$$MnO_4^- + 8H^+ \rightarrow Mn^2+ + 4H_2O$$

Mn goes from +7 to +2, meaning it gains 5 electrons, so we need to add those 5 electrons to the equation

$$MnO_4^- +5e^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$$

Now that this half-reaction is balanced, we can move on to the oxidation half-reaction

$$Fe^{2+} \rightarrow Fe^3+$$

Fe loses an electron in this reaction, so we can add that electron to the right side

$$Fe^{2+} \rightarrow Fe^{3+} + e^-$$

When we combine these reactions, we want the electrons to cancel, so we have to multiply the whole equation by 5.

$$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$$

Lastly, we can combine these two reactions to get the final redox reaction.

$$MnO_4^- + 8H^+ +5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$$

Balance the redox reaction (in basic solution) given the skeleton equation:

$$Zn + NO_3^- \rightarrow Zn^{2+} + NO_2$$

The oxidation states for each species are the same as their charge.

Now onto the half-reactions.

$$Zn \rightarrow Zn^{2+}$$

Zn is being oxidized, since its oxidation number is increasing. This also means we need to add our 2 electrons to the right side.

$$Zn \rightarrow Zn^2+ + 2e^-$$

Now for our reduction reaction,

$$NO_3^- \rightarrow NO_2$$

Next, we add 2H⁺ to the left to balance H

$$NO_3^- + 2H^+ \rightarrow NO_2 + H_2O$$

Since this is a reduction, we add an electron to the left side

$$NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O$$

For our electrons to cancel, we multiply this reaction by 2

$$2NO_3^- + 4H^+ + 2e^- \rightarrow 2NO_2 + 2H_2O$$

Now we can combine our reactions

$$2NO_3 + 4H^+ + Zn \rightarrow Zn^{2+} + 2NO_2 + 2H_2O$$

Now for the fun part. We need to add 4OH^- to both sides, so we can make water to cancel our hydrogen molecules

$$2NO_3 + 4H^+ + Zn + 4OH^- \rightarrow Zn^{2+} + 2NO_2 + 2H_2O + 4OH^- $$

We combine the H+ and OH^- to make water

$$2NO_3 + Zn + 4H_2O \rightarrow Zn^{2+} + 2NO_2 + 2H_2O + 4OH^- $$

Lastly, we can subtract the 2 water molecules on the right from the left. So, our final reaction is:

$$2NO_3 + Zn + 2H_2O \rightarrow Zn^{2+} + 2NO_2 + 4OH^- $$

Which species are oxidized and reduced in this reaction?

$$2H_2O + 2NADP^+ 3ADP + 3P + light \rightarrow 2NADPH + 2H^+ + 3ATP + O_2$$

First things first, we have a bunch of things that, while they are a part of the reaction as a whole, aren't directly a part of the redox. We can simplify this reaction to be:

$$2H_2O + 2NADP^+ \rightarrow 2NADPH + 2H^+ + O_2$$

Now on to giving each species its oxidation number. H has an oxdation state of +1 on both sides, O starts with -2, then changes to 0. NADP is a bit trickier. Instead of calculating the oxidation state, we can look at the half-reaction.

$$NADP^+ \rightarrow NADPH$$$$NADP^+ + H^+ \rightarrow NADPH$$

$$NADP^+ + H^+ + 2e^- \rightarrow NADPH$$

We see that 2e^- are added on the left side to balance the reaction. Since NADP⁺ is gaining electrons, it is being reduced. Oxygen goes from an oxidation state of -2 to 0, so it is being oxidized.