Redox Reactions

In basic medium, $\mathrm{KMnO_4}$ is reduced to $\mathrm{MnO_2}$ and $\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$.

Step 1: Balance the redox reaction in basic medium

Half-reactions:

Reduction:

$ \mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-} $

Oxidation:

$ \mathrm{2I^- \rightarrow I_2 + 2e^-} $

LCM of electrons $=6$:

$ 2(\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}) $

$ 3(\mathrm{2I^- \rightarrow I_2 + 2e^-}) $

Adding:

$ \mathrm{2MnO_4^- + 4H_2O + 6I^- \rightarrow 2MnO_2 + 8OH^- + 3I_2} $

So,

• $2$ mol $\mathrm{MnO_4^-}$ produce $3$ mol $\mathrm{I_2}$.

Step 2: Calculate moles of reactants

Moles of KI:

$ n(\mathrm{I^-})=0.5\times 1.2=0.6\ \text{mol} $

Moles of $\mathrm{KMnO_4}$:

$ n(\mathrm{MnO_4^-})=0.5\times 0.2=0.1\ \text{mol} $

From the balanced equation:

$ 1\ \text{mol }\mathrm{MnO_4^-} \text{ requires } 3\ \text{mol }\mathrm{I^-} $

So $0.1$ mol $\mathrm{MnO_4^-}$ requires $0.3$ mol $\mathrm{I^-}$, which is available ($0.6$ mol).

Hence, $\mathrm{KMnO_4}$ is the limiting reagent.

Step 3: Moles of $\mathrm{I_2}$ liberated

$ n(\mathrm{I_2})=\frac{3}{2}\times 0.1=0.15\ \text{mol} $

Step 4: Titration with thiosulfate

Reaction:

$ \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}} $

So,

$ n(\mathrm{S_2O_3^{2-}})=2\times 0.15=0.30\ \text{mol} $

Given $\mathrm{Na_2S_2O_3}$ is $0.1\ \mathrm{M}$:

$ V=\frac{n}{M}=\frac{0.30}{0.1}=3.0\ \text{L} $

Answer (nearest integer)

$ \boxed{3} $

$ \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O} $

In this reaction, $6 \mathrm{I}^{-}$ ions get oxidised to $3 \mathrm{I}_2$.

Each $\mathrm{I}^{-}$ loses $1$ electron during oxidation, so total electrons lost (in moles) $=\mathrm{X}=6$.

$ \mathrm{Cr}_2 \mathrm{O}_7^{2-}+3 \mathrm{~S}^{2-}+14 \mathrm{H}^{+} \rightarrow+\mathrm{S}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} $

Here, $3 \mathrm{S}^{2-}$ ions get oxidised to $3\mathrm{S}$.

Each $\mathrm{S}^{2-}$ loses $2$ electrons to become $\mathrm{S}$, so total electrons lost (in moles) $=\mathrm{Y}=3 \times 2=6$.

Therefore, $x+y=6+6=12$.

In acidic medium, dichromate oxidises $Fe^{2+}$ to $Fe^{3+}$.

Half-reactions (NCERT method):

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

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

So, $1$ mole of $Cr_2O_7^{2-}$ oxidises $6$ moles of $Fe^{2+}$.

Step 1: Moles of $Fe^{2+}$ in Mohr’s salt solution

Volume $=750\ \text{cc}=0.75\ \text{L}$, Molarity $=0.6\ \text{M}$

$n(Fe^{2+}) = 0.75 \times 0.6 = 0.45\ \text{mol}$

Step 2: Moles of $K_2Cr_2O_7$ required

$n(K_2Cr_2O_7)=\frac{0.45}{6}=0.075\ \text{mol}$

Step 3: Use given dichromate solution data

Volume $=200\ \text{cc}=0.2\ \text{L}$, Molarity $=x\times 10^{-3}\ \text{M}$

$n = 0.2 \times x\times 10^{-3} = 2\times 10^{-4}x$

Equate moles:

$2\times 10^{-4}x = 0.075$

$x=\frac{0.075}{2\times 10^{-4}}=375$

$ \boxed{x = 375} $

$\begin{aligned} & 3 \mathrm{e}^{-}+\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{\Theta} \\ & 2 \mathrm{I}^{\ominus} \rightarrow \mathrm{I}_2+2 \mathrm{e}^{-} \\ & \mathrm{I}_2+2 \mathrm{~S}_2 \mathrm{O}_3^{-2} \rightarrow 2 \mathrm{I}^{\ominus}+\mathrm{S}_4 \mathrm{O}_6^{-2}\end{aligned}$

gm equivalent of $\mathrm{MnO}_4^{-}$

$ =0.2 \times \frac{500}{1000} \times 3=0.3 \text { (Limiting reagent) } $

gm equivalent of KI

$ =1.5 \times \frac{500}{1000} \times 1=0.75 $

gm equivalent of $\mathrm{MnO}_4^{-}$

= gm equivalent of hypo

$ \begin{aligned} & 0.3=x \times \frac{300}{1000} \times 1 \\ & x=1 \end{aligned} $

X is difference in oxidation state.

$7-2=5$

So $X=5$

$\begin{aligned} & 6 \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O} \\ & \rightarrow\left[\mathrm{Fe}_3\left(\mathrm{OH}_2\right)\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+2 \mathrm{H}^{\oplus} \\ & {\left[\mathrm{Fe}_3(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+4 \mathrm{H}_2 \mathrm{O}} \\ & \rightarrow \underset{\text { Brown red ppt }}{\left[\mathrm{Fe}(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right]\right.}+\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}^{\oplus} \end{aligned}$

$\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0$ contains 5 d electrons

So $Y=5$

$X+Y=5+5=10$

Let moles of $\mathrm{CO}_2=\mathrm{n}$

moles of $\mathrm{Ca}(\mathrm{OH})_2$

total initially $=2 \mathrm{n}$

excess $\mathrm{Ca}(\mathrm{OH})_2=\mathrm{n}$

gm equivalent of $\mathrm{Ca}(\mathrm{OH})_2=$ gm equivalent of HCl

$ \begin{aligned} & \mathrm{n} \times 2=0.1 \times \frac{40}{1000} \times 1 \\ & \mathrm{n}=2 \times 10^{-3} \end{aligned} $

Volume of $\mathrm{CO}_2=2 \times 10^{-3} \times 22400=44.8 \mathrm{~cm}^3$

$\begin{aligned} & \mathrm{(M) \times(2) \times(5)=2 \times 20 \times 2} \\ & \mathrm{M=8} \end{aligned} $

$\begin{aligned} & \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7(\mathrm{~s})+4 \mathrm{KCl}(\mathrm{s})+6 \mathrm{H}_2 \mathrm{SO}_4 \text { (conc.) } \\ & \rightarrow 2 \mathrm{CrO}_2 \mathrm{Cl}_2(\mathrm{~g})+6 \mathrm{KHSO}_4+3 \mathrm{H}_2 \mathrm{O} \end{aligned}$

This reaction is called chromyl chloride test.

Here oxidation state of $\mathrm{Cr}$ is +6.

$2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}$

$\therefore$ Number of moles of $\mathrm{H}^{+}$ ions required by 1 mole of $\mathrm{MnO}_4^{-}$ to oxidise oxalate ion to $\mathrm{CO}_2$ is 8

Intermediate oxidation state of element can undergo disproportionation.

$\mathrm{H}_2 \mathrm{O}_2, \mathrm{ClO}_3^{-}, \mathrm{P}_4, \mathrm{Cl}_2, \mathrm{Cu}^{+1}, \mathrm{NO}_2$

Adding oxidation half and reduction half, net reaction is

$\begin{aligned} & 2 \mathrm{MnO}_4^{-}+6 \mathrm{I}^{-}+4 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{I}_2+2 \mathrm{MnO}_2+8 \mathrm{OH}^{-} \\ \Rightarrow \quad & \mathrm{z}=8 \\ \Rightarrow \quad & \text { Ans } 8 \end{aligned}$

$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+} \text {, }$

Oxidation no. of $\mathrm{Fe}=+1$

In alkaline medium, the reduction of permanganate anion involves a gain of 3 electrons.

The balanced chemical equation for the reduction of permanganate anion in alkaline medium is:

In this reaction, the permanganate anion (MnO₄^-) is reduced to manganese dioxide (MnO₂) by gaining 3 electrons. The water molecules (H₂O) are oxidized by losing 2 electrons. The hydroxide ions (OH^-) are spectator ions that do not participate in the redox reaction.

The reduction of permanganate anion in alkaline medium is a useful reaction for a variety of analytical and synthetic purposes. For example, it can be used to quantitatively determine the concentration of an analyte in a solution, or to synthesize a variety of organic compounds.

To find the sum of the oxidation states of the metals in $\mathrm{Fe(CO)}_{5}$, $\mathrm{VO}^{2+}$, and $\mathrm{WO}_{3}$, we first need to find the oxidation states of the metals in each compound.

1. $\mathrm{Fe(CO)}_{5}$: The compound $\mathrm{CO}$ is a neutral ligand, so it does not contribute to the oxidation state of the metal. Hence, the oxidation state of Fe in $\mathrm{Fe(CO)}_{5}$ is 0.




2. $\mathrm{VO}^{2+}$: The compound $\mathrm{O}$ has an oxidation state of -2, and since the overall charge of the compound is +2, the oxidation state of V must be +4 to balance this out.




3. $\mathrm{WO}_{3}$: The compound $\mathrm{O}$ has an oxidation state of -2, and since there are three oxygen atoms, the total oxidation state contributed by oxygen is -6. To balance this out, the oxidation state of W must be +6.

Therefore, the sum of the oxidation states of the metals in these three compounds is $0 + 4 + 6 = 10$.

The sum of the oxidation states of bromine in bromic acid (HBrO₃) and perbromic acid (HBrO₄) is +12.

In bromic acid, the oxidation state of bromine is +5, while in perbromic acid, it is +7. The sum of the oxidation states of bromine in both compounds is (+5) + (+7) = +12.

$\mathop {2Cl{O_2}}\limits_{Chlorine\,dioxide} + 2{O_3}\buildrel {} \over \longrightarrow \mathop {C{l_2}{O_6}}\limits_{Chloral\,perchlorate} + 2{O_2}$

Oxidation state of Cl in Cl₂O₆ = 2x + 6($-$2) = 0

2x $-$ 12 = 0

2x = 12 $\Rightarrow$ x = + 6

Average oxidation state of Cl in Cl₂O₆ is 6.

In neutral or faintly alkaline solution, thiosulphate is oxidized to sulphate by permanganate,

8MnO₄^– + 3S₂O₃^2– + H₂O $ \to $ 8MnO₂ + 6SO₄^2– + 2OH^–

(i) The structure of compound containing sulphur in $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ is :

Let the oxidation state of sulpher be $x$ :

$ \begin{gathered} 4 \times x+2 \times(+1)+6 \times(-2)=0 \\ 4 x=12-2=0 \\ 4 x=10 \\ x=\frac{5}{2}=2.5 \end{gathered} $

(ii) Each of the corner sulphurs utilises five valence electrons to form bond with oxygen atoms.

Their oxidation state is +5.

(iii) Oxidation state of central sulphur atom is zero (0).

Difference between two types of sulphur = 5 – 0 = 5.

When arranging the species I₂, HI, HIO₄, and ICl by the oxidation number of iodine (I) in increasing order, we need to identify the oxidation number of iodine in each compound or molecule.

• In molecular iodine, I₂, iodine is in its elemental form, so its oxidation number is 0.
• In hydrogen iodide, HI, hydrogen usually has an oxidation number of +1 (except in metal hydrides, which is not the case here), and since the compound is neutral, iodine must have an oxidation number of -1 to balance the +1 from hydrogen.
• In iodine monochloride, ICl, chlorine usually has an oxidation state of -1 (except in compounds with oxygen or fluorine). Given that the compound is neutral, iodine must have an oxidation number of +1 to balance the -1 from chlorine.
• In periodic acid, HIO₄, oxygen has a typical oxidation state of -2 (except in peroxides, superoxides, or when bonded to fluorine). Considering there are 4 oxygen atoms, this gives a total oxidation state of -8 from oxygen. Hydrogen, as usual, is +1. To make the compound neutral, the oxidation state of iodine must be such that when added to the +1 from hydrogen and -8 from oxygen, the total is 0. Therefore,

  $+1 + (\text{Oxidation state of I}) - 8 = 0$

  Solving for the oxidation state of iodine:

  $ \text{Oxidation state of I} = 7$

So, the oxidation numbers of iodine in the given species are:

• I₂: 0
• HI: -1
• ICl: +1
• HIO₄: +7

Arranging them in increasing order of the oxidation number of iodine, we get:

HI (-1) < I₂ (0) < ICl (+1) < HIO₄ (+7)

To complete and balance the given equation, it's important to first understand that this is a redox reaction where chlorate ion ($ClO_3^-$) is reduced to chloride ion ($Cl^-$), and iodide ion ($\text{I}^-$) is oxidized to iodine ($I_2$). The presence of sulfuric acid ($\text{H}_2\text{SO}_4$) suggests it mainly acts as a source of protons (H⁺) and the bisulfate ion ($HSO_4^-$) is a byproduct of the reaction.

First, let's write down all the half-reactions and balance them:

$\text{I}^- \to \text{I}_2$

To balance the iodine atoms, we need 2 I^- on the left:

$2\text{I}^- \to \text{I}_2$

$\text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \to \text{Cl}^- + 3\text{H}_2\text{O}$

The iodine half-reaction needs to be balanced in terms of electrons to conclude the overall redox reaction:

$2\text{I}^- \to \text{I}_2 + 2\text{e}^-$

Now, to combine these half-reactions keeping the electron count balanced, we notice that the reduction half-reaction involves 6 electrons, whereas the oxidation half-reaction involves only 2 electrons. Therefore, to balance electrons, we need to multiply the oxidation half-reaction by 3:

$6\text{I}^- \to 3\text{I}_2 + 6\text{e}^-$

Now, adding the balanced half-reactions:

$6\text{I}^- + \text{ClO}_3^- + 6\text{H}^+ \to 3\text{I}_2 + \text{Cl}^- + 3\text{H}_2\text{O}$

Since the reaction occurs in acidic medium (indicated by the presence of sulfuric acid), the $6\text{H}^+$ comes from $3\text{H}_2\text{SO}_4$, but because sulfate ions ($SO_4^{2-}$) are not changed in the reduction process, for every $\text{H}_2\text{SO}_4$ that donates two $H^+$, one $HSO_4^-$ is formed:

$6\text{I}^- + \text{ClO}_3^- + 3\text{H}_2\text{SO}_4 \to 3\text{I}_2 + \text{Cl}^- + 3\text{H}_2\text{O} + 3HSO_4^-$

However, we need to ensure the sulfate ions are properly accounted for. Since each $\text{H}_2\text{SO}_4$ provides two hydrogens and we use 3 $\text{H}_2\text{SO}_4$, it's correct as shown. Thus, the complete and balanced equation in acidic solution is:

$6\text{I}^- + \text{ClO}_3^- + 3\text{H}_2\text{SO}_4 \to 3\text{I}_2 + \text{Cl}^- + 3\text{H}_2\text{O} + 3\text{HSO}_4^-$

To complete and balance the reaction between $Mn^{2+}$ and $PbO_2$ that results in $MnO_4^-$ and $H_2O$, we need to take into account both the redox aspect of this reaction and the need for balancing atoms and charges. This reaction occurs in acidic solution, so we'll be using $H^+$ ions to balance hydrogen atoms and electrons to balance the charges.

The half-reactions are as follows:

Oxidation half-reaction: The manganese goes from a +2 oxidation state in $Mn^{2+}$ to a +7 oxidation state in $MnO_4^-$.

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

We add 4 $H_2O$ on the right side to balance the oxygen, giving 8 $H^+$ on the left side to balance the hydrogen. Then, we add electrons to balance the charge:

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

Reduction half-reaction: The lead goes from a +4 oxidation state in $PbO_2$ to a +2 oxidation state in the product not mentioned explicitly (since $PbO_2$ is being reduced, we can assume it becomes $Pb^{2+}$ in acidic solution).

$PbO_2 + 4H^+ + 2e^- \to Pb^{2+} + 2H_2O$

Now, to balance the electrons in both half-reactions, we need to equalize the number of electrons. Since the oxidation half-reaction produces 5 electrons and the reduction half-reaction consumes 2 electrons, we multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2 to get 10 electrons in both:

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

$5(PbO_2 + 4H^+ + 2e^- \to Pb^{2+} + 2H_2O)$

By multiplying, we have:

$2Mn^{2+} + 16H^+ + 8H_2O \to 2MnO_4^- + 10e^-$

$5PbO_2 + 20H^+ + 10e^- \to 5Pb^{2+} + 10H_2O$

Combining and simplifying these equations, while also canceling out electrons and what's common on both sides, we find:

$2Mn^{2+} + 5PbO_2 + 16H^+ \to 2MnO_4^- + 5Pb^{2+} + 8H_2O$

In this balanced equation, both the charges and the atoms are balanced, giving a clear representation of the chemical reaction that occurs.

To complete and balance the given chemical equation, we must ensure that the number of atoms for each element on the reactant side is the same as on the product side. The given chemical reaction involves nitric acid (HNO₃) reacting with hydrochloric acid (HCl) to produce nitric oxide (NO) and chlorine gas (Cl₂). This reaction is part of the aqua regia process, which is used to dissolve noble metals, and involves redox reactions where the nitrate ion acts as an oxidizing agent.

The initial unbalanced equation is:

$ \text{HNO}_3 + \text{HCl} \rightarrow \text{NO} + \text{Cl}_2 $

To balance the equation, follow these steps:

1. Balance the atoms of elements that appear in only one reactant and one product first. In this reaction, balancing chlorine (Cl) is a good starting point. Notice that there are 2 chlorine atoms on the product side because of Cl₂, so we need 2 HCl molecules on the reactant side to balance the chlorine atoms.
2. Balance the number of nitrogen atoms next. There is 1 nitrogen atom on each side of the equation, so the nitric acid and nitric oxide are already balanced for nitrogen.
3. Balance the oxygen atoms. There is 1 oxygen atom in HNO₃ and none in the products. We must account for this discrepancy. Here, realizing that water (H₂O) is often a product in reactions involving acids can help. Adding water to the products will allow us to balance oxygen.
4. Adjust hydrogen atoms last. After adding water to the products for oxygen balance, we will add hydrogen atoms accordingly to both sides to maintain balance.

Considering the points above, let's add water to the products and then balance the hydrogen atoms:

$ 3\text{HNO}_3 + 6\text{HCl} \rightarrow \text{NO} + 3\text{Cl}_2 + 2\text{H}_2\text{O} $

This equation is now balanced with respect to all atoms:

• Nitrogen: There are 3 nitrogen atoms on the reactant side and 1 nitrogen atom on the product side.
• Oxygen: There are 3 oxygen atoms from HNO₃ on the reactant side and 2 from water (H₂O) on the product side.
• Chlorine: There are 6 chlorine atoms from HCl on the reactant side and 3$\times$2=6 chlorine atoms in Cl₂ on the product side.
• Hydrogen: There are 6 hydrogen atoms from HCl on the reactant side and 2$\times$2=4 hydrogen atoms in 2 molecules of water (H₂O).

This reflects the complete balanced chemical equation for the reaction involving nitric acid and hydrochloric acid to produce nitric oxide, chlorine gas, and water:

$ 3\text{HNO}_3 + 6\text{HCl} \rightarrow \text{NO} + 3\text{Cl}_2 + 2\text{H}_2\text{O} $

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 $NH_4^+$.