Redox Reactions

To complete and balance the given redox reaction between cerium(III) ions ($Ce^{3+}$) and peroxodisulfate ions ($S_2O_8^{2-}$) to form sulfate ions ($SO_4^{2-}$) and cerium(IV) ions ($Ce^{4+}$), we need to follow a step-by-step approach. This involves balancing the atoms and charges through oxidation and reduction half-reactions.

First, we separate the reaction into half-reactions:

Oxidation half-reaction (cerium is oxidized):

$Ce^{3+} \to Ce^{4+} + e^-$

Reduction half-reaction (peroxodisulfate is reduced):

$S_2O_8^{2-} + 2e^- \to 2SO_4^{2-}$

Next, we balance the electrons between the two half-reactions. The oxidation half-reaction releases 1 electron, while the reduction half-reaction consumes 2 electrons. To balance the electrons, we multiply the oxidation half-reaction by 2:

Oxidation half-reaction balanced with electrons:

$2Ce^{3+} \to 2Ce^{4+} + 2e^-$

Now, the electrons in both half-reactions are balanced, and we can combine them:

$2Ce^{3+} + S_2O_8^{2-} \to 2Ce^{4+} + 2SO_4^{2-}$

Finally, we have a balanced redox reaction. Both the number of atoms and the charges are balanced:

$2Ce^{3+} + S_2O_8^{2-} \to 2Ce^{4+} + 2SO_4^{2-}$

This reaction shows that two cerium(III) ions are oxidized to cerium(IV) ions, while a peroxodisulfate ion is reduced to two sulfate ions, with the transfer of two electrons to maintain charge balance.

To complete and balance the reaction between chlorine gas (Cl₂) and hydroxide ions (OH^-) to produce chloride ions (Cl^-) and hypochlorite ions (ClO^-), we need to take into account both the conservation of mass and the conservation of charge. This reaction occurs in aqueous solution, and it depends on the conditions such as temperature and the concentration of the hydroxide ions. There are actually two possible reactions. In a basic solution, we can have a balanced equation that reflects the production of both chloride and hypochlorite ions:

$\text{Cl}_2 + 2\text{OH}^- \to \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}$

Explanation:

- The chlorine gas (Cl₂) reacts with hydroxide ions (OH^-) present in the aqueous solution.

- Two hydroxide ions are required to balance the equation, ensuring that the chlorine from Cl₂ is distributed between the produced chloride ions (Cl^-) and hypochlorite ions (ClO^-).

- The equation is balanced as written because there are two chlorine atoms on each side, the charges are balanced (2 negative charges on the reactant side from the hydroxide ions and 2 negative charges on the product side from Cl^- and ClO^- ions), and the additional product is water (H₂O), formed by the combination of the hydrogen from the hydroxide ions and an oxygen atom from one of the hydroxide ions.

This equation highlights how the presence of OH^- drives the disproportionation of Cl₂, splitting it into two different oxidation states in the products (Cl^- at -1 and ClO^- at +1). This is a common type of reaction for halogens in the presence of base.

To complete and balance the given equation, which involves the dichromate ion ($Cr_2O_7^{2-}$) reacting with ethylene glycol (C₂H₄O) to form acetic acid (C₂H₄O₂) and chromium (III) ion (Cr³⁺), we need to follow a systematic approach. First, let's write down the unbalanced equation:

$Cr_2O_7^{2-} + C_2H_4O \to C_2H_4O_2 + Cr^{3+}$

Next, we balance the equation in acidic solution using the half-reaction method, which involves balancing the reduction and oxidation half-reactions separately and then combining them.

Oxidation (Ethylene glycol to acetic acid):

$C_2H_4O \to C_2H_4O_2$

Reduction (Dichromate to Chromium (III)):

$Cr_2O_7^{2-} \to Cr^{3+}$

Oxidation doesn’t need balancing in this respect. For the reduction:

$Cr_2O_7^{2-} \to 2Cr^{3+}$

For the reduction half:

$Cr_2O_7^{2-} \to 2Cr^{3+} + 7H_2O$

For the reduction half:

$Cr_2O_7^{2-} + 14H^+ \to 2Cr^{3+} + 7H_2O$

The oxidation half does not involve oxygen or hydrogen atoms changing, so we skip balancing them directly here.

Oxidation half (2 electrons are needed to balance the change from +2 to +3 oxidation state for 2 Carbon atoms):

$C_2H_4O \to C_2H_4O_2 + 2e^-$

Reduction half (to balance it, we need to add 6 electrons):

$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$

To balance electrons, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1 (since 6 electrons are transferred in the reduction and 2 in the oxidation; the least common multiple is 6).

$3(C_2H_4O \to C_2H_4O_2 + 2e^-)$

$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$

Combine and simplify:

$Cr_2O_7^{2-} + 14H^+ + 3C_2H_4O \to 3C_2H_4O_2 + 2Cr^{3+} + 7H_2O$

$Cr_2O_7^{2-} + 14H^+ + 3C_2H_4O \to 3C_2H_4O_2 + 2Cr^{3+} + 7H_2O$

This represents the balanced chemical equation for the reaction between $Cr_2O_7^{2-}$ and ethylene glycol in an acidic medium, producing acetic acid and chromium (III) ions alongside water.

To complete and balance the given chemical equation:

$Zn + NO_3^- \to Zn^{2+} + NH_4^+$

We start by looking at the oxidation and reduction processes separately. In this reaction:

• Zinc (Zn) is oxidized to zinc ions ($Zn^{2+}$).
• The nitrate ion ($NO_3^-$) is reduced to ammonium ions ($NH_4^+$).

To balance this equation, we should keep in mind that both mass and charge must be conserved. Therefore, we have to balance not only the atoms but also the charges throughout the equation.

First, let's write the half-reactions:

• Oxidation (loss of electrons): $Zn \to Zn^{2+} + 2e^-$
• Reduction (gain of electrons): Since nitrate ($NO_3^-$) is reducing to ammonia ($NH_4^+$), let's assume a general reduction half-reaction that will be balanced later: $NO_3^- + xH^+ + ye^- \to NH_4^+ + zH_2O$

Now, balancing the reduction half-reaction involves getting the nitrogen from a +5 oxidation state in $NO_3^-$ to a -3 state in $NH_4^+$. This process consumes 8 electrons (e-) to reduce each nitrogen atom, also hydrogen ions ($H^+$) are required to form $NH_4^+$, and water ($H_2O$) is a product due to the oxygen from the nitrate.

The correct balanced reduction half-reaction, considering water as the source of $H^+$ and the formation of $H_2O$, is:

$NO_3^- + 10H^+ + 8e^- \to NH_4^+ + 3H_2O$

Thus, combining our half-reactions and ensuring the electrons are balanced:

• For the zinc oxidation: $Zn \to Zn^{2+} + 2e^-$
• For the reduction of nitrate: $NO_3^- + 10H^+ + 8e^- \to NH_4^+ + 3H_2O$

Since the reduction half-reaction consumes 8 electrons and the oxidation produces only 2, we need to multiply the oxidation half-reaction by 4 to balance the electrons:

$4Zn \to 4Zn^{2+} + 8e^-$

The full reaction, adding the two half-reactions together and balancing mass and charge, now looks like this:

$4Zn + 10H^+ + NO_3^- \to 4Zn^{2+} + NH_4^+ + 3H_2O$

It is important to note that this equation assumes acidic conditions due to the presence of $H^+$ ions. However, the original equation did not specify the reaction environment, so this balanced equation is based on typical conditions that allow for $NO_3^-$ reduction to $NH_4^+$.