Chapter 7: Redox Reactions (Redox Reactions)

(a) $1+2+x-8=0$: P = +5. (b) $1+1+x-8=0$: S = +6. (c) $4+2x-14=0$: P = +5. (d) $2+x-8=0$: Mn = +6. (e) CaO₂ is a peroxide (Ca²⁺O₂²⁻): O = −1. (f) H is −1 (hydride): $1+x-4=0$: B = +3. (g) $2+2x-14=0$: S = +6. (h) water is neutral: $1+3+2(x-8)=0$: S = +6.

(a) +5 (b) +6 (c) +5 (d) +6 (e) −1 (f) +3 (g) +6 (h) +6

(a) Average −1/3. KI₃ is K⁺[I–I←I⁻]: two I atoms are 0 (in I₂) and one is −1.

(b) Average +2.5. The two terminal S atoms (bonded to three O) are +5; the two central S (S–S bonds) are 0: (5 + 0 + 0 + 5)/4 = 2.5.

(c) Average +8/3: Fe₃O₄ = FeO·Fe₂O₃ has one Fe²⁺ and two Fe³⁺.

(d) Average −2: the CH₃ carbon is −3, the CH₂OH carbon −1.

(e) Average 0: the CH₃ carbon is −3, the COOH carbon +3.

Fractional values are averages; chemically different atoms of the same element have different, whole-number oxidation states.

(a) −1/3 (b) +2.5 (c) +8/3 (d) −2 (e) 0 (averages of different whole-number states)

In each, one element's oxidation number rises and another's falls.

(a) Cu +2 → 0 (reduced); H 0 → +1 (oxidised). (b) Fe +3 → 0 (reduced); C +2 → +4 (oxidised). (c) Taking H as +1 in B₂H₆ (as NCERT does): B +3 → −3 (reduced); H −1 (in LiAlH₄) → +1 (oxidised). (d) K 0 → +1 (oxidised); F 0 → −1 (reduced). (e) N −3 → +2 (oxidised); O 0 → −2 (reduced).

Each involves simultaneous oxidation and reduction.

Oxidation numbers: in H₂O, O = −2; in F₂, F = 0; in HF, F = −1; in HOF, H = +1 and F = −1, so O = 0.

Fluorine is reduced (0 → −1) and oxygen is oxidised (−2 → 0).

F₂ is reduced and the oxygen of water is oxidised: a redox reaction.

By formula: H₂SO₅: $2+x-10=0\Rightarrow x=+8$, impossible (S has only 6 valence electrons). Its structure has a peroxide link, HO–O–S(=O)₂–OH: two O atoms are −1, so $2+x-6-2=0\Rightarrow$ S = +6.

Cr₂O₇²⁻: $2x-14=-2\Rightarrow$ Cr = +6 (two CrO₄ tetrahedra sharing one O; no peroxide): correct.

NO₃⁻: $x-6=-1\Rightarrow$ N = +5 (planar, one N=O and two N–O⁻ in resonance): correct.

The fallacy (+8 for S) arises only when peroxide oxygen is wrongly counted as −2.

S +6 (not +8, because of a peroxide link), Cr +6, N +5

(a) HgCl₂ (b) NiSO₄ (c) SnO₂ (d) Tl₂SO₄ (e) Fe₂(SO₄)₃ (f) Cr₂O₃

Carbon: −4 CH₄; −3 C₂H₆; −2 CH₃Cl (or CH₃OH); −1 C₂H₂; 0 CH₂Cl₂ (or HCHO); +1 (CHO)₂; +2 CO (or CHCl₃); +3 (COOH)₂; +4 CO₂ (or CCl₄).

Nitrogen: −3 NH₃; −2 N₂H₄; −1 NH₂OH; 0 N₂; +1 N₂O; +2 NO; +3 HNO₂ (N₂O₃); +4 NO₂; +5 HNO₃ (N₂O₅).

As listed above

S in SO₂ is +4 (its range is −2 to +6) and O in H₂O₂ is −1 (range −2 to 0): both are in intermediate states and can go up (reductant) or down (oxidant). N in HNO₃ is at its highest state (+5) and can only be reduced; ozone (O = 0) can only be reduced to −1 or −2 in its reactions. So both act only as oxidants.

SO₂ and H₂O₂ are in intermediate oxidation states; HNO₃ and O₃ can only be reduced.

(a) In photosynthesis all the O₂ comes from water, not from CO₂: 12 H₂O give 6 O₂ (the H reduces CO₂), and 6 new H₂O are formed from the oxygen of CO₂.

(b) One O₂ comes from O₃ (O₃ → O₂ + O) and the other from H₂O₂ (oxidised by that O atom), so they are written separately.

Technique: isotopic labelling with ¹⁸O (H₂¹⁸O or H₂¹⁸O₂) and tracing where the labelled oxygen appears (mass spectrometry).

They show the true source of each O₂; use ¹⁸O tracer studies.

In AgF₂ silver is +2, an unstable state for Ag (stable state +1, 4d¹⁰). Ag²⁺ readily takes an electron (Ag²⁺ + e⁻ → Ag⁺), oxidising other substances.

Ag²⁺ readily gains an electron to become the stable Ag⁺.

(i) $\mathrm{P}{}_4+6\mathrm{Cl}{}_2\to 4\mathrm{PCl}{}_3$ (P in excess, P = +3); $\mathrm{P}{}_4+10\mathrm{Cl}{}_2\to 4\mathrm{PCl}{}_5$ (Cl₂ in excess, P = +5).

(ii) $2\mathrm{C}+\mathrm{O}{}_2\to 2\mathrm{CO}$ (C in excess, +2); $\mathrm{C}+\mathrm{O}{}_2\to \mathrm{CO}{}_2$ (O₂ in excess, +4).

(iii) $4\mathrm{Na}+\mathrm{O}{}_2\to 2\mathrm{Na}{}_2\mathrm{O}$ (Na in excess, O = −2); $2\mathrm{Na}+\mathrm{O}{}_2\to \mathrm{Na}{}_2\mathrm{O}{}_2$ (O₂ in excess, O = −1).

PCl₃/PCl₅, CO/CO₂, Na₂O/Na₂O₂

(a) Toluene does not mix with aqueous KMnO₄; in alcohol both dissolve, giving a homogeneous mixture and a faster reaction. The medium becomes alkaline, so benzoate forms, which is acidified to benzoic acid:

$\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{CH}{}_3+2\mathrm{MnO}{}_4{}^{-}\to \mathrm{C}{}_6\mathrm{H}{}_5\mathrm{COO}{}^{-}+2\mathrm{MnO}{}_2+\mathrm{OH}{}^{-}+\mathrm{H}{}_2\mathrm{O}$

(b) $2\mathrm{NaCl}+\mathrm{H}{}_2\mathrm{SO}{}_4\to 2\mathrm{HCl}+\mathrm{Na}{}_2\mathrm{SO}{}_4$: HCl is too weak a reductant to reduce H₂SO₄, so it escapes as a colourless gas. With bromide, HBr is formed first, but HBr is a stronger reductant and reduces H₂SO₄: $2\mathrm{HBr}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Br}{}_2+\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}$, giving red-brown bromine vapour.

(a) Alcohol dissolves both reactants (b) HBr reduces H₂SO₄ to give Br₂; HCl cannot.

As in the table

Average oxidation state of S: S₂O₃²⁻ +2, S₄O₆²⁻ +2.5, SO₄²⁻ +6. I₂ is a weak oxidant and raises S only to +2.5 (tetrathionate); Br₂ is a stronger oxidant (E° 1.09 V vs 0.54 V) and takes S to +6 (sulphate).

Br₂ is a stronger oxidant than I₂.

F₂ displaces every other halogen from its halides: $\mathrm{F}{}_2+2\mathrm{Cl}{}^{-}\to 2\mathrm{F}{}^{-}+\mathrm{Cl}{}_2$, $\mathrm{F}{}_2+2\mathrm{Br}{}^{-}\to 2\mathrm{F}{}^{-}+\mathrm{Br}{}_2$ (highest E°, +2.87 V); no halogen can oxidise F⁻.

HI is the best reductant: it reduces conc. H₂SO₄ to H₂S ($8\mathrm{HI}+\mathrm{H}{}_2\mathrm{SO}{}_4\to 4\mathrm{I}{}_2+\mathrm{H}{}_2\mathrm{S}+4\mathrm{H}{}_2\mathrm{O}$) and Cu²⁺ to Cu⁺ ($2\mathrm{Cu}{}^{2+}+4\mathrm{I}{}^{-}\to \mathrm{Cu}{}_2\mathrm{I}{}_2+\mathrm{I}{}_2$), which HCl cannot do. Reducing power: HF < HCl < HBr < HI.

F₂ oxidises all halides; HI reduces what the other acids cannot.

Xe is reduced from +8 to +6 while F⁻ is oxidised to F₂. Since F₂ is the strongest elemental oxidant, only an even stronger oxidant can oxidise F⁻.

Na₄XeO₆ (perxenate) is a stronger oxidising agent than F₂.

H₃PO₂ (a strong reductant) reduces both Ag⁺ and Cu²⁺ to the metals. The weaker reductant benzaldehyde reduces Ag⁺ (Tollens' test) but not Cu²⁺.

Both act as oxidants, but Ag⁺ is a stronger oxidant than Cu²⁺ (E° 0.80 V vs 0.34 V).

(a) $2\mathrm{I}{}^{-}\to \mathrm{I}{}_2+2\mathrm{e}{}^{-}$ (×3); $\mathrm{MnO}{}_4{}^{-}+2\mathrm{H}{}_2\mathrm{O}+3\mathrm{e}{}^{-}\to \mathrm{MnO}{}_2+4\mathrm{OH}{}^{-}$ (×2):

$2\mathrm{MnO}{}_4{}^{-}+6\mathrm{I}{}^{-}+4\mathrm{H}{}_2\mathrm{O}\to 2\mathrm{MnO}{}_2+3\mathrm{I}{}_2+8\mathrm{OH}{}^{-}$

(b) $\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}\to \mathrm{HSO}{}_4{}^{-}+3\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}$ (×5); $\mathrm{MnO}{}_4{}^{-}+8\mathrm{H}{}^{+}+5\mathrm{e}{}^{-}\to \mathrm{Mn}{}^{2+}+4\mathrm{H}{}_2\mathrm{O}$ (×2):

$2\mathrm{MnO}{}_4{}^{-}+5\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}+\mathrm{H}{}^{+}\to 2\mathrm{Mn}{}^{2+}+5\mathrm{HSO}{}_4{}^{-}$

(c) $\mathrm{Fe}{}^{2+}\to \mathrm{Fe}{}^{3+}+\mathrm{e}{}^{-}$ (×2); $\mathrm{H}{}_2\mathrm{O}{}_2+2\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}\to 2\mathrm{H}{}_2\mathrm{O}$:

$\mathrm{H}{}_2\mathrm{O}{}_2+2\mathrm{Fe}{}^{2+}+2\mathrm{H}{}^{+}\to 2\mathrm{Fe}{}^{3+}+2\mathrm{H}{}_2\mathrm{O}$

(d) $\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}\to \mathrm{SO}{}_4{}^{2-}+4\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}$ (×3); $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+14\mathrm{H}{}^{+}+6\mathrm{e}{}^{-}\to 2\mathrm{Cr}{}^{3+}+7\mathrm{H}{}_2\mathrm{O}$:

$\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+3\mathrm{SO}{}_2+2\mathrm{H}{}^{+}\to 2\mathrm{Cr}{}^{3+}+3\mathrm{SO}{}_4{}^{2-}+\mathrm{H}{}_2\mathrm{O}$

As balanced above

(a) The product is hypophosphite, H₂PO₂⁻ (P = +1). P goes 0 → −3 (PH₃) and 0 → +1: one P falls by 3 while three P rise by 1 each.

$\mathrm{P}{}_4+3\mathrm{OH}{}^{-}+3\mathrm{H}{}_2\mathrm{O}\to \mathrm{PH}{}_3+3\mathrm{H}{}_2\mathrm{PO}{}_2{}^{-}$. P₄ is both oxidant and reductant (disproportionation).

(b) N: −2 → +2 (8 e⁻ per N₂H₄); Cl: +5 → −1 (6 e⁻): ratio 3 N₂H₄ : 4 ClO₃⁻.

$3\mathrm{N}{}_2\mathrm{H}{}_4+4\mathrm{ClO}{}_3{}^{-}\to 6\mathrm{NO}+4\mathrm{Cl}{}^{-}+6\mathrm{H}{}_2\mathrm{O}$. Oxidant ClO₃⁻, reductant N₂H₄.

(c) Cl: +7 → +3 (8 e⁻ per Cl₂O₇); O: −1 → 0 (2 e⁻ per H₂O₂): ratio 1 : 4.

$\mathrm{Cl}{}_2\mathrm{O}{}_7+4\mathrm{H}{}_2\mathrm{O}{}_2+2\mathrm{OH}{}^{-}\to 2\mathrm{ClO}{}_2{}^{-}+4\mathrm{O}{}_2+5\mathrm{H}{}_2\mathrm{O}$. Oxidant Cl₂O₇, reductant H₂O₂.

(Both methods give the same equations.)

As balanced above

Oxidation state of C (taking N as −3): (CN)₂ +3, CN⁻ +2, CNO⁻ +4. Cyanogen is reduced and oxidised at the same time: a disproportionation reaction. Cyanogen behaves like a halogen (a pseudohalogen), cf. $\mathrm{Cl}{}_2+2\mathrm{OH}{}^{-}\to \mathrm{Cl}{}^{-}+\mathrm{ClO}{}^{-}+\mathrm{H}{}_2\mathrm{O}$.

A disproportionation; cyanogen is a pseudohalogen.

$\mathrm{Mn}{}^{3+}+\mathrm{e}{}^{-}\to \mathrm{Mn}{}^{2+}$; $\mathrm{Mn}{}^{3+}+2\mathrm{H}{}_2\mathrm{O}\to \mathrm{MnO}{}_2+4\mathrm{H}{}^{+}+\mathrm{e}{}^{-}$. Adding:

$2\mathrm{Mn}{}^{3+}+2\mathrm{H}{}_2\mathrm{O}\to \mathrm{Mn}{}^{2+}+\mathrm{MnO}{}_2+4\mathrm{H}{}^{+}$

(a) F (most electronegative: only −1). (b) Cs (only +1). (c) I (−1 to +7). (d) Ne (forms no compounds).

(a) F (b) Cs (c) I (d) Ne

Cl₂ is reduced to Cl⁻ and SO₂ is oxidised to SO₄²⁻.

$\mathrm{Cl}{}_2+\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}\to \mathrm{H}{}_2\mathrm{SO}{}_4+2\mathrm{HCl}$

Disproportionation needs an element with at least three oxidation states.

(a) P, S, Cl, Br, I (e.g. P₄, S₈, Cl₂ in alkali). (b) Cu ($2\mathrm{Cu}{}^{+}\to \mathrm{Cu}+\mathrm{Cu}{}^{2+}$), Mn (Mn³⁺, MnO₄²⁻), Ga (also In).

(a) P, S, Cl, Br, I (b) Cu, Mn, Ga

$4\mathrm{NH}{}_3+5\mathrm{O}{}_2\to 4\mathrm{NO}+6\mathrm{H}{}_2\mathrm{O}$. $n(\mathrm{NH}{}_3)=\frac{10.00}{17}=0.588$ mol needs $\frac{5}{4}\times 0.588=0.735$ mol O₂, but only $\frac{20.00}{32}=0.625$ mol is available: O₂ is limiting.

NO $=0.625\times \frac{4}{5}=0.500$ mol $\times 30$ g mol⁻¹.

15.0 g of NO

$E_{\text{cell}}^{\circ }=E^{\circ }(\text{species reduced})-E^{\circ }(\text{species oxidised})$:

(a) 0.77 − 0.54 = +0.23 V (b) 0.80 − 0.34 = +0.46 V (c) 0.77 − 0.34 = +0.43 V (d) 0.77 − 0.80 = −0.03 V (e) 1.09 − 0.77 = +0.32 V

Feasible: (a), (b), (c), (e); not feasible: (d)

(i) Cathode: $\mathrm{Ag}{}^{+}+\mathrm{e}{}^{-}\to \mathrm{Ag}$ (deposited); anode: $\mathrm{Ag}\to \mathrm{Ag}{}^{+}+\mathrm{e}{}^{-}$ (the anode dissolves).

(ii) Cathode: Ag deposited; inert anode: $2\mathrm{H}{}_2\mathrm{O}\to \mathrm{O}{}_2+4\mathrm{H}{}^{+}+4\mathrm{e}{}^{-}$ (O₂; HNO₃ forms).

(iii) Cathode: H₂; anode: O₂ (water is electrolysed).

(iv) Cathode: Cu; anode: Cl₂.

(i) Ag deposits, anode dissolves (ii) Ag and O₂ (iii) H₂ and O₂ (iv) Cu and Cl₂

A metal displaces any metal below it in the activity (electrochemical) series.

Mg > Al > Zn > Fe > Cu

The more negative the E°, the stronger the reducing agent.

Ag < Hg < Cr < Mg < K

Cell: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s).

(i) The zinc electrode (anode) is negative. (ii) Electrons carry the current in the external wire (from Zn to Ag); ions carry it in the solutions and through the salt bridge. (iii) Anode: $\mathrm{Zn}\to \mathrm{Zn}{}^{2+}+2\mathrm{e}{}^{-}$ (oxidation); cathode: $\mathrm{Ag}{}^{+}+\mathrm{e}{}^{-}\to \mathrm{Ag}$ (reduction). $E_{\text{cell}}^{\circ }=0.80-(-0.76)=1.56$ V.

Zn | Zn²⁺ || Ag⁺ | Ag; Zn is negative; electrons outside, ions inside