Redox Reactions

Using the rules for oxidation number:

- In **NaH$_2$PO$_4$, the underlined element is P**.
Let oxidation number of P be $x$.

$+1 + 2(+1) + x + 4(-2) = 0$

$1 + 2 + x - 8 = 0 \Rightarrow x = +5$

- In **NaHSO$_4$, the underlined element is S**.

$+1 + 1 + x + 4(-2) = 0$

$2 + x - 8 = 0 \Rightarrow x = +6$

- In **H$_4$P$_2$O$_7$, the underlined element is P**.

$4(+1) + 2x + 7(-2) = 0$

$4 + 2x - 14 = 0 \Rightarrow 2x = 10 \Rightarrow x = +5$

- In **K$_2$MnO$_4$, the underlined element is Mn**.

$2(+1) + x + 4(-2) = 0$

$2 + x - 8 = 0 \Rightarrow x = +6$

- In **CaO$_2$, it is a peroxide, so each O has oxidation number $-1$.

- In NaBH$_4$, the underlined element is B**.
In borohydride, H is $-1$.

$+1 + x + 4(-1) = 0$

$1 + x - 4 = 0 \Rightarrow x = +3$

- In **H$_2$S$_2$O$_7$, the underlined element is S**.

$2(+1) + 2x + 7(-2) = 0$

$2 + 2x - 14 = 0 \Rightarrow 2x = 12 \Rightarrow x = +6$

- In **KAl(SO$_4$)$_2\cdot$12H$_2$O, the underlined element is Al.
Aluminium has oxidation number $+3$ in its compounds.

So the oxidation numbers are:
P = +5, S = +6, P = +5, Mn = +6, O = -1, B = +3, S = +6, Al = +3**.

For each species, first assign the oxidation number and then explain the result.

- **KI$_3$: The iodine here is effectively present as I$_2$ + I$^-$, so the average oxidation number of iodine is $-\tfrac{1}{3}$. This is a case of fractional oxidation number, which is only an average value.

- H$_2$S$_4$O$_6$**: Let the average oxidation number of sulphur be $x$.

$2(+1) + 4x + 6(-2) = 0$

$2 + 4x - 12 = 0 \Rightarrow 4x = 10 \Rightarrow x = +2.5$

The real structure has sulphur atoms in different oxidation states, not a true fractional state for each atom.

- **Fe$_3$O$_4$**: Oxygen is $-2$. So total for oxygen is $4(-2)=-8$.

Let average oxidation number of Fe be $x$.

$3x - 8 = 0 \Rightarrow x = +\tfrac{8}{3}$.

This is a mixed oxide; structurally it contains Fe in +2 and +3 states.

- **CH$_3$CH$_2$OH: The underlined carbon atoms have different oxidation numbers.
- Carbon in CH$_3$**: three C–H bonds give $-3$, so oxidation number is **$-3$.
- Carbon in CH$_2$OH**: two C–H bonds give $-2$, one C–O bond gives $+1$, so oxidation number is **$-1$.

- CH$_3$COOH: Again, the two carbon atoms differ.
- Carbon in CH$_3$: oxidation number $-3$.
- Carbon in COOH**: one C=O contributes $+2$, one C–O contributes $+1$, so oxidation number is **$+3$.

These results show that fractional oxidation numbers are average values. The actual structure reveals that atoms of the same element may exist in different whole-number oxidation states**.

Assign oxidation numbers and show the changes.

(a) $\mathrm{CuO(s)} + \mathrm{H}_2(g) \rightarrow \mathrm{Cu(s)} + \mathrm{H}_2\mathrm{O(g)}$

- Cu in CuO is +2 and becomes 0 in Cu, so Cu is reduced.
- H in H$_2$ is 0 and becomes +1 in H$_2$O, so H$_2$ is oxidised.

Therefore, it is a redox reaction.

(b) $\mathrm{Fe}_2\mathrm{O}_3(s) + 3\mathrm{CO}(g) \rightarrow 2\mathrm{Fe}(s) + 3\mathrm{CO}_2(g)$

- Fe in Fe$_2$O$_3$ is +3 and becomes 0 in Fe, so Fe is reduced.
- C in CO is +2 and becomes +4 in CO$_2$, so CO is oxidised.

Therefore, it is a redox reaction.

(c) $4\mathrm{BCl}_3(g) + 3\mathrm{LiAlH}_4(s) \rightarrow 2\mathrm{B}_2\mathrm{H}_6(g) + 3\mathrm{LiCl}(s) + 3\mathrm{AlCl}_3(s)$

- In BCl$_3$, B is +3; in B$_2$H$_6$, B is negative/less positive than +3, so boron is reduced.
- In LiAlH$_4$, hydride acts as reducing agent and Al is carried into AlCl$_3$ where Al is +3; effectively, the hydride ion transfers electrons and is oxidised.

So oxidation and reduction occur simultaneously, hence this is a redox reaction.

(d) $2\mathrm{K}(s) + \mathrm{F}_2(g) \rightarrow 2\mathrm{K}^+\mathrm{F}^-(s)$

- K goes from 0 to +1: oxidised.
- F goes from 0 to -1: reduced.

Hence, this is a redox reaction.

(e) $4\mathrm{NH}_3(g) + 5\mathrm{O}_2(g) \rightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_2\mathrm{O}(g)$

- N in NH$_3$ is -3 and in NO is +2, so nitrogen is oxidised.
- O in O$_2$ is 0 and in NO/H$_2$O is -2, so oxygen is reduced.

Therefore, this is a redox reaction.

For the reaction

$\mathrm{H}_2\mathrm{O}(s) + \mathrm{F}_2(g) \rightarrow \mathrm{HF}(g) + \mathrm{HOF}(g)$

assign oxidation numbers:

- In $\mathrm{F}_2$, fluorine is 0.
- In HF, F is -1.
- In HOF, H is +1, O is -2, so F is +1.

Thus fluorine in $\mathrm{F}_2$ is both reduced and oxidised:
- one F atom changes from 0 to -1 in HF, so it is reduced;
- the other F atom changes from 0 to +1 in HOF, so it is oxidised.

Hence the reaction is a disproportionation redox reaction.

Find the oxidation numbers and then relate them to the structures.

- **H$_2$SO$_5$**: Let oxidation number of S be $x$.

This is a peroxide-type compound because one O–O bond is present, so the peroxide oxygens are $-1$ and the rest are $-2$.

Using the formula:

$2(+1) + x + 5(-2)$ would not fit the real structure if all O were taken as $-2$.

In the actual structure, sulphur is +6.

- **Cr$_2$O$_7^{2-}$**: Let oxidation number of Cr be $x$.

$2x + 7(-2) = -2$

$2x - 14 = -2 \Rightarrow 2x = 12 \Rightarrow x = +6$

- **NO$_3^-$**: Let oxidation number of N be $x$.

$x + 3(-2) = -1$

$x - 6 = -1 \Rightarrow x = +5$

Structures and the fallacy of fractional oxidation state:

- In **H$_2$SO$_5$, one oxygen is present as a peroxide bond** ($-1$ oxidation state for those oxygens), while the remaining oxygens are $-2$. So the compound is better represented by its structure, not by a single uniform oxidation number for all oxygens.
- In **Cr$_2$O$_7^{2-}** and **NO$_3^-$, the oxidation numbers found are whole numbers and consistent with the structure.

The main point is that oxidation number is a formal bookkeeping tool; the real bonding must be understood from the structure**.

Use the given oxidation states of the metals:

- Mercury(II) chloride → **HgCl$_2$
- Nickel(II) sulphate → NiSO$_4$
- Tin(IV) oxide → SnO$_2$
- Thallium(I) sulphate → Tl$_2$SO$_4$
- Iron(III) sulphate → Fe$_2$(SO$_4$)$_3$
- Chromium(III) oxide → Cr$_2$O$_3$**

So the correct set is the first option.

We need examples showing the range of oxidation states.

For carbon:
- in **CH$_4$, C is -4
- in CO, C is +2
- in CO$_2$, C is +4
- in CCl$_4$, C is also +4

For nitrogen:
- in NH$_3$, N is -3
- in N$_2$H$_4$, N is -2
- in NO, N is +2
- in NO$_2$, N is +4
- in HNO$_3$, N is +5**

So one valid list is the one given in the correct option.

Sulphur dioxide and hydrogen peroxide can act as both oxidising and reducing agents because the elements involved, sulphur in SO$_2$ and oxygen in H$_2$O$_2$, are in intermediate oxidation states.

- In **SO$_2$, sulphur is in the +4 oxidation state. It can be oxidised to +6 or reduced** to 0 or -2, so SO$_2$ can act as both oxidant and reductant.
- In **H$_2$O$_2$, oxygen is in the -1 oxidation state. It can be oxidised** to 0 in O$_2$ or reduced to -2 in H$_2$O, so H$_2$O$_2$ also acts as both.

Ozone and nitric acid act only as oxidants because:
- In **O$_3$, oxygen is in the 0 state and generally gets reduced to -2. It cannot readily be oxidised further.
- In HNO$_3$, nitrogen is in its highest oxidation state, +5. It can only be reduced, not oxidised further.

Hence ozone and nitric acid behave only as oxidising agents**.

The reactions show that the same element appears in different oxidation states and one element is both oxidised and reduced.

- (a) $6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2$
- Carbon in CO$_2$ is +4 and is reduced in glucose.
- Oxygen in water is -2 and becomes 0 in O$_2$.
- The more proper form is written with water on both sides so that the source and fate of oxygen are clear:

$6\mathrm{CO}_2 + 12\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{H}_2\mathrm{O} + 6\mathrm{O}_2$

- (b) $\mathrm{O}_3 + \mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} + 2\mathrm{O}_2$
- O in ozone and peroxide is rearranged; one oxygen is reduced and another is oxidised.
- The more proper form makes the two oxygen molecules explicit:

$\mathrm{O}_3 + \mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{O}_2 + \mathrm{O}_2$

Technique to investigate the path of the reactions:
- Use isotopic labelling of oxygen, such as $^{18}$O, to trace which oxygen atoms come from which reactant.

The compound **AgF$_2$ contains silver in the +2 oxidation state**, which is much less stable than Ag(I). Since silver prefers the lower oxidation state, Ag(II) readily gets reduced to Ag(I) by accepting electrons.

Therefore, if AgF$_2$ is formed, it acts as a very strong oxidising agent because it can easily oxidise other substances while itself getting reduced.

This agrees with the fact that compounds in higher, less stable oxidation states are usually strong oxidants.

This happens because the product oxidation state depends on the relative amounts of oxidant and reductant available.

- If the reducing agent is in excess, the oxidant is not fully pushed to its highest possible oxidation state, so a lower oxidation state product forms.
- If the oxidising agent is in excess, the reducing agent is oxidised more completely, so a higher oxidation state product forms.

Illustrations:

1. Phosphorus with alkali

With excess alkali, phosphorus gives phosphine and hypophosphite:

$\mathrm{P}_4 + 3\mathrm{OH}^- + 3\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{PH}_3 + 3\mathrm{H}_2\mathrm{PO}_2^-$

Here phosphorus ends up in lower and higher oxidation states depending on conditions.

2. Nitrogen dioxide in alkali

$2\mathrm{NO}_2 + 2\mathrm{OH}^- \rightarrow \mathrm{NO}_2^- + \mathrm{NO}_3^- + \mathrm{H}_2\mathrm{O}$

Nitrogen in NO$_2$ (+4) disproportionates to +3 in NO$_2^-$ and +5 in NO$_3^-$.

3. Chlorine in alkali

$\mathrm{Cl}_2 + 2\mathrm{OH}^- \rightarrow \mathrm{Cl}^- + \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O}$

Chlorine goes to a lower oxidation state (-1) and a higher oxidation state (+1).

So the final oxidation state depends on which reagent is in excess.

- (a) Concentrated sulphuric acid is not a strong enough oxidant to oxidise chloride ion further, so chloride gives HCl gas, which is colourless and pungent.
- (b) Bromide ion is more easily oxidised than chloride ion, so concentrated sulphuric acid oxidises bromide to **Br$_2$, giving red vapours.

Thus, the difference is due to the reducing strength of halide ions**:
- $\mathrm{Cl}^-$ is harder to oxidise, so mainly HCl is liberated.
- $\mathrm{Br}^-$ is easier to oxidise, so Br$_2$ is formed.

This is consistent with the order of reducing power of halide ions: **I$^-$ > Br$^-$ > Cl$^-$**.

Assign oxidation number changes in each reaction.

(a) $2\mathrm{AgBr} + \mathrm{C}_6\mathrm{H}_6\mathrm{O}_2 \rightarrow 2\mathrm{Ag} + 2\mathrm{HBr} + \mathrm{C}_6\mathrm{H}_4\mathrm{O}_2$
- AgBr is reduced to Ag.
- The organic compound $\mathrm{C}_6\mathrm{H}_6\mathrm{O}_2$ is oxidised to $\mathrm{C}_6\mathrm{H}_4\mathrm{O}_2$.
- Oxidising agent: AgBr
- Reducing agent: $\mathrm{C}_6\mathrm{H}_6\mathrm{O}_2$

(b) $\mathrm{HCHO} + 2[\mathrm{Ag(NH}_3)_2]^+ + 3\mathrm{OH}^- \rightarrow 2\mathrm{Ag} + \mathrm{HCOO}^- + 4\mathrm{NH}_3 + 2\mathrm{H}_2\mathrm{O}$
- Carbon in HCHO is oxidised to formate.
- $[\mathrm{Ag(NH}_3)_2]^+$ is reduced to Ag.
- Oxidising agent: $[\mathrm{Ag(NH}_3)_2]^+$
- Reducing agent: HCHO

(c) $\mathrm{HCHO} + 2\mathrm{Cu}^{2+} + 5\mathrm{OH}^- \rightarrow \mathrm{Cu}_2\mathrm{O} + \mathrm{HCOO}^- + 3\mathrm{H}_2\mathrm{O}$
- Carbon in HCHO is oxidised to formate.
- Cu$^{2+}$ is reduced to Cu$_2$O.
- Oxidising agent: Cu$^{2+}$
- Reducing agent: HCHO

(d) $\mathrm{N}_2\mathrm{H}_4 + 2\mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O}$
- Nitrogen in N$_2$H$_4$ is oxidised to N$_2$.
- Oxygen in H$_2$O$_2$ is reduced to H$_2$O.
- Oxidising agent: H$_2$O$_2$
- Reducing agent: N$_2$H$_4$

(e) $\mathrm{Pb} + \mathrm{PbO}_2 + 2\mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{PbSO}_4 + 2\mathrm{H}_2\mathrm{O}$
- Pb(0) is oxidised to Pb$^{2+}$ in PbSO$_4$.
- Pb in PbO$_2$ is reduced from +4 to +2.
- Oxidising agent: PbO$_2$
- Reducing agent: Pb

Compare the role of thiosulphate in the two reactions.

1. With iodine:

$2\mathrm{S}_2\mathrm{O}_3^{2-} + \mathrm{I}_2 \rightarrow \mathrm{S}_4\mathrm{O}_6^{2-} + 2\mathrm{I}^-$

Thiosulphate is oxidised only to tetrathionate. So iodine is a relatively mild oxidising agent.

2. With bromine:

$\mathrm{S}_2\mathrm{O}_3^{2-} + 2\mathrm{Br}_2 + 5\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{SO}_4^{2-} + 4\mathrm{Br}^- + 10\mathrm{H}^+$

Thiosulphate is oxidised much further to sulphate. So bromine is a much stronger oxidising agent than iodine.

Therefore, the same reductant reacts differently because bromine has greater oxidising power than iodine and can oxidise thiosulphate more completely.

Among halogens, fluorine is the best oxidant because it has the highest tendency to gain electrons.

This is shown by its reactions:
- It can displace other halogens from halides.
- It can even oxidise water:

$2\mathrm{H}_2\mathrm{O} + 2\mathrm{F}_2 \rightarrow 4\mathrm{HF} + \mathrm{O}_2$

This shows fluorine is stronger than all other halogens as an oxidising agent.

Among hydrohalic acids, hydroiodic acid (HI) is the best reductant because iodide ion is most easily oxidised.

The reducing power of halides follows:
**I$^-$ > Br$^-$ > Cl$^-$ > F$^-$**

So HI can be oxidised most readily, which makes it the strongest reducing hydrohalic acid.