Redox Reactions

(a) $NaH_2\underline{P}O_4$: $1(+1) + 2(+1) + x + 4(-2) = 0 \Rightarrow 3 + x – 8 = 0 \Rightarrow x = \mathbf{+5}$.

(b) $NaH\underline{S}O_4$: $1 + 1 + x + 4(-2) = 0 \Rightarrow 2 + x – 8 = 0 \Rightarrow x = \mathbf{+6}$.

(c) $H_4\underline{P}_2O_7$: $4(+1) + 2x + 7(-2) = 0 \Rightarrow 4 + 2x – 14 = 0 \Rightarrow 2x = 10 \Rightarrow x = \mathbf{+5}$.

(d) $K_2\underline{Mn}O_4$: $2(+1) + x + 4(-2) = 0 \Rightarrow 2 + x – 8 = 0 \Rightarrow x = \mathbf{+6}$.

(e) $Ca\underline{O}_2$: (Peroxide) $2 + 2x = 0 \Rightarrow x = \mathbf{-1}$.

(f) $Na\underline{B}H_4$: (Hydride) $1 + x + 4(-1) = 0 \Rightarrow 1 + x – 4 = 0 \Rightarrow x = \mathbf{+3}$.

(g) $H_2\underline{S}_2O_7$: $2 + 2x – 14 = 0 \Rightarrow 2x = 12 \Rightarrow x = \mathbf{+6}$.

(h) $KAl(\underline{S}O_4)_2 \cdot 12H_2O$: K(+1), Al(+3). Water is neutral. Total charge of sulfate is -2. $x + 4(-2) = -2 \Rightarrow x – 8 = -2 \Rightarrow x = \mathbf{+6}$.

(a) $K\underline{I}_3$: Average O.N. $= -1/3$. Structure: $K^+ [I – I \leftarrow I]^-$. A complex ion $I_3^-$.

(b) $H_2\underline{S}_4O_6$: Average $= 2.5$. Structure ($O_3S-S-S-SO_3$ part). Terminal S are +5, Middle two S are 0. Avg: $(5+0+0+5)/4 = 2.5$.

(c) $Fe_3O_4$: Mixed oxide ($FeO \cdot Fe_2O_3$). Fe has +2 and +3 states. Avg = 8/3.

(d) $\underline{C}H_3\underline{C}H_2OH$: $C_1$ (attached to 3H, 1C) has -3. $C_2$ (attached to 2H, 1O, 1C) has -1.

(e) $\underline{C}H_3\underline{C}OOH$: Methyl C is -3. Carboxyl C (+3 due to 2 bonds to O, 1 to OH). Avg 0.

A reaction is redox if oxidation numbers of elements change.

(a) $Cu(+2) \to Cu(0)$ (Reduction). $H(0) \to H(+1)$ (Oxidation).

(b) $Fe(+3) \to Fe(0)$ (Red). $C(+2) \to C(+4)$ (Ox).

(c) $B(+3) \to B(+3)$. But $H(-1)$ in $LiAlH_4$ goes to $H(-1)$ in $B_2H_6$. However, this is a redox reaction where Al acts as reducing agent transferring Hydride. Formally, H remains -1, but chemical environment changes significantly.

(d) $K(0) \to K(+1)$ (Ox). $F(0) \to F(-1)$ (Red).

(e) $N(-3) \to N(+2)$ (Ox). $O(0) \to O(-2)$ (Red).

$F_2(0) \to F(-1)$ in HF (Reduction).
$F_2(0) \to F(+1)$ in HOF? No, Fluorine is always -1 in compounds. In HOF, F is -1, H is +1, so Oxygen must be 0.
Therefore, $O(-2)$ in water $\to O(0)$ in HOF (Oxidation).
Thus, it is a redox reaction.

$H_2SO_5$ (Caro’s Acid): Calc $x$: $2 + x – 10 = 0 \Rightarrow x=8$ (Wrong, max +6).
Structure has Peroxide linkage (-O-O-). Two O are -1, Three O are -2.
$2(+1) + x + 2(-1) + 3(-2) = 0 \Rightarrow x – 6 = 0 \Rightarrow x = \mathbf{+6}$.

$Cr_2O_7^{2-}$: Structure has Cr-O-Cr. All O are -2. No peroxide.
$2x + 7(-2) = -2 \Rightarrow 2x = 12 \Rightarrow x = \mathbf{+6}$.

$NO_3^-$: Calc: $x – 6 = -1 \Rightarrow x = \mathbf{+5}$. Max valency of N is 4, but Oxidation State can be +5 (formal charge).

(a) Mercury(II) chloride: $\mathbf{HgCl_2}$
(b) Nickel(II) sulphate: $\mathbf{NiSO_4}$
(c) Tin(IV) oxide: $\mathbf{SnO_2}$
(d) Thallium(I) sulphate: $\mathbf{Tl_2SO_4}$
(e) Iron(III) sulphate: $\mathbf{Fe_2(SO_4)_3}$
(f) Chromium(III) oxide: $\mathbf{Cr_2O_3}$

Carbon (-4 to +4):
-4: $CH_4$
-2: $CH_3Cl$ (approx), $C_2H_4$
0: $C$ (Diamond/Graphite), $CH_2Cl_2$
+2: $CO$, $CHCl_3$
+4: $CO_2$, $CCl_4$

Nitrogen (-3 to +5):
-3: $NH_3$
-2: $N_2H_4$
-1: $NH_2OH$
0: $N_2$
+1: $N_2O$
+2: $NO$
+3: $N_2O_3$
+4: $NO_2$
+5: $N_2O_5, HNO_3$

$SO_2$: Sulphur is in +4 state. Range of S is -2 to +6. Since +4 is intermediate, it can increase to +6 (reducing agent) or decrease to 0/-2 (oxidising agent).
$O_3$: Oxygen is 0. It readily gives nascent oxygen, reducing to -2 (most stable). It acts only as oxidant.
$HNO_3$: Nitrogen is in +5 state (Max). It can only decrease. Hence, acts only as oxidant.

(a) Photosynthesis: $6CO_2 + 12H_2O \to C_6H_{12}O_6 + 6H_2O + 6O_2$.
This emphasizes that all $O_2$ evolved comes from water, not $CO_2$. Technique: Use radioisotope $^{18}O$ in water.
(b) Ozone with H₂O₂: $O_3 + H_2O_2 \to H_2O + O_2 + O_2$.
One molecule of oxygen comes from ozone and the other from hydrogen peroxide. Technique: Use labelled $^{18}O$ in one reactant.

In $AgF_2$, Silver is in +2 state. This is unstable as Ag is essentially stable in +1 state ($4d^{10}$). Thus, $Ag^{2+}$ readily accepts an electron to become $Ag^+$, acting as a very strong oxidising agent.

If reducing agent is in excess, lower oxidation state product forms. If oxidising agent is in excess, higher oxidation state forms.

• $C + O_2 (\text{limited}) \to CO$ (+2). $C + O_2 (\text{excess}) \to CO_2$ (+4).
• $P_4 + Cl_2 (\text{limited}) \to PCl_3$ (+3). $P_4 + Cl_2 (\text{excess}) \to PCl_5$ (+5).
• $Na + O_2 (\text{limited}) \to Na_2O$ (-2). $Na + O_2 (\text{excess}) \to Na_2O_2$ (-1, peroxo).

(a) Alcoholic KMnO4: Toluene is organic (non-polar). Ionic KMnO4 is insoluble. Alcohol acts as a solvent allowing reactants to mix. The reaction yields Benzoic acid.

(b) H2SO4 with Bromide: HBr formed is a strong reducing agent. Conc. H2SO4 is an oxidising agent. It oxidises HBr to $Br_2$ (red vapour). With Chloride, HCl is formed. HCl is not reducing enough to be oxidised by H2SO4 to Cl2. Hence, HCl gas evolves.

(a) $AgBr$ (Oxidant), Hydroquinone (Reductant).

(b) $HCHO$ (Reductant, oxidised to HCOO-), $[Ag(NH_3)_2]^+$ (Oxidant, reduced to Ag).

(c) $HCHO$ (Reductant), $Cu^{2+}$ (Oxidant).

(d) $N_2H_4$ (Reductant, oxidised to N2), $H_2O_2$ (Oxidant, reduced to H2O).

(e) $Pb$ (Reductant, 0 to +2), $PbO_2$ (Oxidant, +4 to +2). This is a comproportionation reaction.

Bromine ($Br_2$) is a stronger oxidising agent than Iodine ($I_2$).
Bromine oxidises Sulphur from +2 (in $S_2O_3^{2-}$) all the way to +6 (in $SO_4^{2-}$).
Iodine is a weaker oxidant; it only oxidises Sulphur from +2 to +2.5 (in $S_4O_6^{2-}$ tetrathionate).

Fluorine (Best Oxidant): High electronegativity and small size allow it to accept electrons easily. Reaction: $F_2 + 2Cl^- \to 2F^- + Cl_2$.

HI (Best Reductant): Iodide ion ($I^-$) has a large size and low bond dissociation energy of H-I, making it easy to lose an electron. Reaction: $2HI + H_2SO_4 \to I_2 + SO_2 + 2H_2O$.

Reaction: $XeO_6^{4-} + F^- \to XeO_3 + F_2$.
Here $F^-$ (-1) is oxidised to $F_2$ (0). This means $XeO_6^{4-}$ acts as an oxidising agent strong enough to oxidise Fluoride ions.
Conclusion: $Na_4XeO_6$ is an extremely powerful oxidising agent.

Reactions show that $Ag^+$ can oxidise both $H_3PO_2$ and Benzaldehyde ($C_6H_5CHO$).
$Cu^{2+}$ can oxidise $H_3PO_2$ but cannot oxidise Benzaldehyde.
Inference: $Ag^+$ is a stronger oxidising agent than $Cu^{2+}$.

(a) Basic: $2MnO_4^- + I^- + H_2O \to 2MnO_2 + IO_3^- + 2OH^-$? No, product is $I_2$.
Balanced: $\mathbf{2MnO_4^- + 6I^- + 4H_2O \to 2MnO_2 + 3I_2 + 8OH^-}$.

(b) Acidic: $\mathbf{2MnO_4^- + 5SO_2 + 2H_2O \to 2Mn^{2+} + 5HSO_4^-}$. (Often $5SO_4^{2-} + 4H^+$).

(c) Acidic: $\mathbf{H_2O_2 + 2Fe^{2+} + 2H^+ \to 2Fe^{3+} + 2H_2O}$.

(d) Acidic: $\mathbf{Cr_2O_7^{2-} + 3SO_2 + 2H^+ \to 2Cr^{3+} + 3SO_4^{2-} + H_2O}$.

(a) $P_4 \to PH_3 + HPO_2^-$: (Disproportionation)
Balanced: $\mathbf{P_4 + 3OH^- + 3H_2O \to PH_3 + 3H_2PO_2^-}$.

(b) $N_2H_4 + ClO_3^- \to NO + Cl^-$:
Balanced: $\mathbf{3N_2H_4 + 4ClO_3^- \to 6NO + 4Cl^- + 6H_2O}$.

(c) $Cl_2O_7 + H_2O_2 \to ClO_2^- + O_2$:
Balanced: $\mathbf{Cl_2O_7 + 4H_2O_2 + 2OH^- \to 2ClO_2^- + 4O_2 + 5H_2O}$.

Reaction: $(CN)_2 + 2OH^- \to CN^- + CNO^- + H_2O$.
This is a disproportionation reaction.
Cyanogen behaves like a halogen (pseudohalogen). Carbon (in CN) goes from oxidation state +3 (in cyanogen) to +2 (in cyanide) and +4 (in cyanate)? Actually, standard calculation is simpler: $(CN)_2$ (0) $\to CN^-$ (-1) and $CNO^-$ (+1). It mimics $Cl_2 + OH^- \to Cl^- + ClO^-$.

Oxidation: $Mn^{3+} \to MnO_2 (+4)$ (Loss of 1e).
Reduction: $Mn^{3+} \to Mn^{2+} (+2)$ (Gain of 1e).
Ratio is 1:1.

(a) Only Negative: F (Most electronegative, always -1).
(b) Only Positive: Cs (Alkali metal, always +1).
(c) Both Pos & Neg: I (Halogen, -1 to +7).
(d) Neither: Ne (Inert gas, usually 0).

Sulphur dioxide reduces Chlorine to Chloride ion, while getting oxidised to Sulphate.
Reaction: $\mathbf{Cl_2(aq) + SO_2(aq) + 2H_2O(l) \to 2Cl^-(aq) + SO_4^{2-}(aq) + 4H^+(aq)}$.

(a) Non-metals: P, S, Cl, Br, I. (Must have intermediate oxidation states).
(b) Metals: Cu (+1 to 0, +2), Mn (+3 to +2, +4), Ga (+2).

Molar Masses: $NH_3 = 17, O_2 = 32, NO = 30$.
Initial: 10g $NH_3$ ($10/17 = 0.588$ mol). 20g $O_2$ ($20/32 = 0.625$ mol).
Stoichiometry: 4 mol NH3 needs 5 mol O2. (Ratio 1.25).
Available Ratio: $0.625 / 0.588 = 1.06$.
Since $1.06 < 1.25$, $O_2$ is limiting.

Calculation: 5 mol $O_2$ gives 4 mol $NO$.
0.625 mol $O_2$ gives $\frac{4}{5} \times 0.625 = 0.5$ mol $NO$.
Mass of NO = $0.5 \times 30 = \mathbf{15.0 \text{ g}}$.

Using $E^\circ_{cell} = E^\circ_{cathode} – E^\circ_{anode}$. If positive, feasible.

(a) $Fe^{3+} + I^- \to Fe^{2+} + I_2$: $E^\circ = 0.77 – 0.54 = +0.23V$. Feasible.

(b) $Ag^+ + Cu \to Ag + Cu^{2+}$: $E^\circ = 0.80 – 0.34 = +0.46V$. Feasible.

(c) $Fe^{3+} + Cu \to Fe^{2+} + Cu^{2+}$: $E^\circ = 0.77 – 0.34 = +0.43V$. Feasible.

(d) $Ag + Fe^{3+} \to Ag^+ + Fe^{2+}$: $E^\circ = 0.77 – 0.80 = -0.03V$. Not Feasible.

(e) $Br_2 + Fe^{2+} \to Br^- + Fe^{3+}$: $E^\circ = 1.09 – 0.77 = +0.32V$. Feasible.