Redox Reactions Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1For a neutral species, the algebraic sum of oxidation numbers is zero. For an ion, it equals the charge on the ion. In ordinary oxides oxygen is −2, while hydrogen is +1 except in metal hydrides.
2. 2(a) In NaH₂PO₄, let the oxidation number of P be x. Na is +1, H is +1 and O is −2:
3. 3
   $$(+1)+2(+1)+x+4(-2)=0\Rightarrow x=+5$$
4. 4Thus the underlined P has oxidation number +5.
5. 5(b) In NaHSO₄, let the oxidation number of S be x:
6. 6
   $$(+1)+(+1)+x+4(-2)=0\Rightarrow x=+6$$
7. 7Thus S has oxidation number +6.
8. 8(c) In H₄P₂O₇, let the oxidation number of P be x:
9. 9
   $$4(+1)+2x+7(-2)=0\Rightarrow x=+5$$
10. 10Each P has oxidation number +5.
11. 11(d) In K₂MnO₄, let the oxidation number of Mn be x:
12. 12
    $$2(+1)+x+4(-2)=0\Rightarrow x=+6$$
13. 13Thus Mn has oxidation number +6.
14. 14(e) CaO₂ is a peroxide, so each O is −1 rather than −2:
15. 15
    $$(+2)+2(-1)=0$$
16. 16Thus the underlined O has oxidation number −1.
17. 17(f) In NaBH₄, H is −1 in the metal hydride and B has oxidation number x:
18. 18
    $$(+1)+x+4(-1)=0\Rightarrow x=+3$$
19. 19Thus B has oxidation number +3.
20. 20(g) In H₂S₂O₇, let the oxidation number of S be x:
21. 21
    $$2(+1)+2x+7(-2)=0\Rightarrow x=+6$$
22. 22Each S has oxidation number +6.
23. 23(h) The twelve waters of crystallisation are neutral. For the sulphate part of KAl(SO₄)₂, let the oxidation number of S be x:
24. 24
    $$(+1)+(+3)+2x+8(-2)=0\Rightarrow x=+6$$
25. 25Thus each S has oxidation number +6.

Step-by-step solution

1. 1The charge-balance equation gives an average value. A fractional average signals that the atoms are not all in the same chemical environment.
2. 2(a) In KI₃, K is +1, so the average oxidation number of iodine is −1/3. The triiodide ion is I⁻–I–I (more precisely I⁻–I₂), giving one I at −1 and two I atoms at 0.
3. 3Thus the average is −1/3, but the individual oxidation states are −1, 0 and 0.
4. 4(b) In H₂S₄O₆, let the average oxidation number of S be x:
5. 5
   $$2(+1)+4x+6(-2)=0\Rightarrow x=+2.5$$
6. 6The tetrathionate structure has two terminal S atoms at +5 and two inner S atoms at 0, so the average is (+5 + 0 + 0 + 5)/4 = +2.5.
7. 7(c) In Fe₃O₄, the average oxidation number of Fe is +8/3. Magnetite is represented as FeO·Fe₂O₃, so one Fe is +2 and two Fe atoms are +3.
8. 8(d) In CH₃CH₂OH, the average carbon oxidation number is −2. The methyl carbon is −3 and the carbon bonded to oxygen is −1.
9. 9(e) In CH₃COOH, the average carbon oxidation number is 0. The methyl carbon is −3 and the carboxyl carbon is +3.
10. 10The important distinction is that an average oxidation number is not necessarily the oxidation number of any individual atom.

Step-by-step solution

1. 1(a) Cu changes from +2 in CuO to 0 in Cu, so it is reduced. H changes from 0 in H₂ to +1 in H₂O, so it is oxidised.
2. 2(b) Fe changes from +3 in Fe₂O₃ to 0 in Fe, so it is reduced. C changes from +2 in CO to +4 in CO₂, so it is oxidised.
3. 3(c) Using the oxidation-number convention intended for this exercise, B changes from +3 in BCl₃ to −3 in B₂H₆, while H changes from −1 in LiAlH₄ to +1 in B₂H₆. BCl₃ is reduced and LiAlH₄ is oxidised.
4. 4(d) K changes from 0 to +1, so K is oxidised. F changes from 0 to −1, so F is reduced.
5. 5(e) N changes from −3 in NH₃ to +2 in NO, so N is oxidised. O changes from 0 in O₂ to −2 in H₂O, so O is reduced.
6. 6In every case at least one oxidation number increases and another decreases; therefore each reaction is redox.

Step-by-step solution

1. 1In F₂, fluorine has oxidation number 0. In HF, F is −1, so one fluorine atom is reduced.
2. 2In HOF, H is +1 and O is −2. Since the molecule is neutral, F is +1, so the other fluorine atom is oxidised.
3. 3The same element in the reactant undergoes both reduction and oxidation, so the reaction is a disproportionation reaction of fluorine.
4. 4
   $${\mathrm{H}}_2\mathrm{O}+{\mathrm{F}}_2⟶\mathrm{HF}+\mathrm{HOF}$$

Step-by-step solution

1. 1H₂SO₅ is peroxymonosulphuric acid with the structure H–O–S(=O)₂–O–O–H. The two O atoms in the O–O peroxide linkage are −1, the other three O atoms are −2, H is +1, and S is +6.
2. 2Check for H₂SO₅: the two H atoms contribute +2, the three ordinary O atoms contribute −6 and the two peroxide O atoms contribute −2. Therefore 2 − 6 − 2 + S = 0, giving S = +6.
3. 3Cr₂O₇²⁻: 2x + 7(−2) = −2, so x = +6 for each Cr. The ion has an O₃Cr–O–CrO₃ structure with two tetrahedral CrO₄ units joined through oxygen.
4. 4NO₃⁻: x + 3(−2) = −1, so x = +5 for N. Nitrate is trigonal planar and resonance-delocalised over the three N–O bonds.
5. 5The fallacy is treating every oxygen as −2. A peroxide O–O bond gives −1 to each oxygen, even in an oxyacid; this is why H₂SO₅ must not be assigned S = +8.

Step-by-step solution

1. 1(a) Hg²⁺ and Cl⁻ combine in a 1:2 ratio, giving HgCl₂.
2. 2(b) Ni²⁺ and SO₄²⁻ combine in a 1:1 ratio, giving NiSO₄.
3. 3(c) Sn⁴⁺ and O²⁻ combine in a 1:2 ratio, giving SnO₂.
4. 4(d) Tl⁺ and SO₄²⁻ combine in a 2:1 ratio, giving Tl₂SO₄.
5. 5(e) Fe³⁺ and SO₄²⁻ combine in a 2:3 ratio, giving Fe₂(SO₄)₃.
6. 6(f) Cr³⁺ and O²⁻ combine in a 2:3 ratio, giving Cr₂O₃.

Step-by-step solution

1. 1For carbon, the complete range is: −4 in CH₄; −3 in C₂H₆; −2 in CH₃OH; −1 in C₂H₂; 0 in CH₂Cl₂; +1 in ClC≡CCl; +2 in CHCl₃ or CO; +3 in CCl₃CCl₃; and +4 in CCl₄ or CO₂.
2. 2For nitrogen, the complete range is: −3 in NH₃; −2 in N₂H₄; −1 in N₂H₂; 0 in N₂; +1 in N₂O; +2 in NO; +3 in N₂O₃; +4 in NO₂; and +5 in N₂O₅.
3. 3These are examples; several different substances can represent the same oxidation state.

Step-by-step solution

1. 1SO₂ contains S at +4. It can be oxidised to +6, acting as a reducing agent, or reduced to lower states, acting as an oxidising agent.
2. 2H₂O₂ contains O at −1. It can be oxidised to O₂ at 0, acting as a reducing agent, or reduced to H₂O with O at −2, acting as an oxidising agent.
3. 3O₃ contains O at 0, its highest common state in the molecule. It accepts electrons and is reduced, so it acts as an oxidant.
4. 4HNO₃ contains N at +5, a high oxidation state. It is readily reduced to lower oxidation states such as +4, +3, +2 or 0, so it acts as an oxidant.
5. 5The general test is whether the element in the substance has a state above, below, or between the accessible states needed for oxidation or reduction.

Step-by-step solution

1. 1(a) Water is both consumed and produced during photosynthesis. The more complete net equation must show the 6H₂O produced, so the reactant amount is 12H₂O:
2. 2
   $$6C{\mathrm{O}}_2+12{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{H}}_2\mathrm{O}+6{\mathrm{O}}_2$$
3. 3Use water labelled with ¹⁸O. The ¹⁸O label appears in the evolved O₂, showing that the released oxygen comes from water.
4. 4(b) The two O₂ molecules have different origins: one comes from O₃ and the other from H₂O₂. A stepwise representation is:
5. 5
   $${\mathrm{O}}_3⟶{\mathrm{O}}_2+\mathrm{O}$$
6. 6
   $${\mathrm{H}}_2{\mathrm{O}}_2+\mathrm{O}⟶{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$$
7. 7Adding the two steps gives O₃ + H₂O₂ → H₂O + O₂ + O₂. Label either O₃ or H₂O₂ with ¹⁸O to identify the source of each oxygen gas molecule; ¹⁸O is a stable tracer.

Step-by-step solution

1. 1In AgF₂, F is −1 and silver is +2.
2. 2Silver in the +2 state readily accepts an electron and returns to its more stable +1 state:
3. 3
   $$\mathrm{A}{\mathrm{g}}^{2+}+{\mathrm{e}}^{-}⟶\mathrm{A}{\mathrm{g}}^{+}$$
4. 4Because Ag²⁺ is strongly electron-accepting, AgF₂ is a powerful oxidising agent, even though it is thermodynamically unstable.

Step-by-step solution

1. 1With phosphorus, excess P₄ limits the fluorinating agent and favours the lower fluoride:
2. 2
   $${\mathrm{P}}_4+6{\mathrm{F}}_2⟶4P{\mathrm{F}}_3$$
3. 3With excess F₂, the higher oxidation state is formed:
4. 4
   $${\mathrm{P}}_4+10{\mathrm{F}}_2⟶4P{\mathrm{F}}_5$$
5. 5With potassium, excess metal forms the lower oxide:
6. 6
   $$4K+{\mathrm{O}}_2⟶2{\mathrm{K}}_2\mathrm{O}$$
7. 7With excess oxygen, potassium peroxide is favoured:
8. 8
   $$2K+{\mathrm{O}}_2⟶{\mathrm{K}}_2{\mathrm{O}}_2$$
9. 9With carbon, excess carbon favours CO:
10. 10
    $$2C+{\mathrm{O}}_2⟶2CO$$
11. 11With excess oxygen, CO₂ is favoured:
12. 12
    $$\mathrm{C}+{\mathrm{O}}_2⟶\mathrm{C}{\mathrm{O}}_2$$
13. 13Thus the relative amount of oxidant or reductant controls how far oxidation or reduction proceeds.

Step-by-step solution

1. 1(a) Alcoholic KMnO₄ provides a convenient organic reaction medium and avoids the need to acidify or strongly alkalinise the mixture. The balanced equation for oxidation of the methyl group is:
2. 2
   $${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_3+2KMn{\mathrm{O}}_4⟶{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOK}+2Mn{\mathrm{O}}_2+\mathrm{KOH}+{\mathrm{H}}_2\mathrm{O}$$
3. 3The methyl carbon changes from −3 to +3, a six-electron oxidation; Mn changes from +7 to +4. The product is potassium benzoate.
4. 4(b) Concentrated H₂SO₄ liberates volatile HCl or HBr from the corresponding halide:
5. 5
   $$2NaCl+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4⟶2NaHS{\mathrm{O}}_4+2HCl\uparrow$$
6. 6
   $$2NaBr+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4⟶2NaHS{\mathrm{O}}_4+2HBr$$
7. 7HBr is a sufficiently strong reducing agent to reduce H₂SO₄, while HCl is not:
8. 8
   $$2HBr+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4⟶\mathrm{B}{\mathrm{r}}_2+\mathrm{S}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$$
9. 9Thus the bromide mixture gives red bromine vapour, whereas the chloride mixture releases colourless, pungent HCl gas.

Step-by-step solution

1. 1(a) C₆H₆O₂ is oxidised to C₆H₄O₂, so it is the reducing agent. Ag⁺ in AgBr is reduced to Ag, so AgBr is the oxidising agent.
2. 2(b) Carbon in HCHO changes from 0 to +2 in HCOO⁻, so HCHO is oxidised and is the reducing agent. Ag⁺ in the diamminesilver(I) complex is reduced to Ag, so the complex is the oxidising agent.
3. 3(c) HCHO is oxidised from carbon 0 to +2, so it is the reducing agent. Cu²⁺ is reduced to Cu⁺ in Cu₂O, so Cu²⁺ is the oxidising agent.
4. 4(d) N changes from −2 in N₂H₄ to 0 in N₂, so N₂H₄ is the reducing agent. O in H₂O₂ changes from −1 to −2, so H₂O₂ is the oxidising agent.
5. 5(e) Pb metal changes from 0 to +2 and is oxidised, so Pb is the reducing agent. Pb in PbO₂ changes from +4 to +2, so PbO₂ is the oxidising agent.
6. 6The substance that loses electrons is oxidised; the species that accepts electrons is the oxidising agent.

Step-by-step solution

1. 1The question concerns (a) iodine and (b) bromine. In thiosulphate, the terminal sulfur is −1 and the central sulfur is +5, giving an average sulfur state of +2.
2. 2With iodine, thiosulphate is oxidised only to tetrathionate. The oxidation and reduction half-reactions are:
3. 3
   $$2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}⟶{\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{e}}^{-}$$
4. 4
   $${\mathrm{I}}_2+2{\mathrm{e}}^{-}⟶2{\mathrm{I}}^{-}$$
5. 5In S₄O₆²⁻, the average sulfur oxidation state is +2.5, so iodine stops at tetrathionate rather than fully oxidising thiosulphate to sulphate.
6. 6With bromine, thiosulphate is oxidised all the way to sulphate:
7. 7
   $${\mathrm{S}}_2{\mathrm{O}}_3^{2-}+5{\mathrm{H}}_2\mathrm{O}⟶2S{\mathrm{O}}_4^{2-}+10{\mathrm{H}}^{+}+8{\mathrm{e}}^{-}$$
8. 8
   $$\mathrm{B}{\mathrm{r}}_2+2{\mathrm{e}}^{-}⟶2B{\mathrm{r}}^{-}$$
9. 9
   $${\mathrm{S}}_2{\mathrm{O}}_3^{2-}+5{\mathrm{H}}_2\mathrm{O}+4B{\mathrm{r}}_2⟶2S{\mathrm{O}}_4^{2-}+8B{\mathrm{r}}^{-}+10{\mathrm{H}}^{+}$$
10. 10Bromine has a higher reduction potential than iodine, so it accepts electrons more readily and oxidises thiosulphate further. In both reactions thiosulphate is the reducing agent and the halogen is the oxidising agent.

Step-by-step solution

1. 1Fluorine has the greatest tendency to gain electrons and can oxidise Cl⁻, Br⁻ and I⁻:
2. 2
   $${\mathrm{F}}_2+2C{\mathrm{l}}^{-}⟶2{\mathrm{F}}^{-}+\mathrm{C}{\mathrm{l}}_2$$
3. 3
   $${\mathrm{F}}_2+2B{\mathrm{r}}^{-}⟶2{\mathrm{F}}^{-}+\mathrm{B}{\mathrm{r}}_2$$
4. 4
   $${\mathrm{F}}_2+2{\mathrm{I}}^{-}⟶2{\mathrm{F}}^{-}+{\mathrm{I}}_2$$
5. 5The reverse displacement by Cl₂, Br₂ or I₂ cannot oxidise F⁻, so the oxidising order is F₂ > Cl₂ > Br₂ > I₂.
6. 6HI and HBr reduce concentrated H₂SO₄, whereas HCl and HF do not under these conditions:
7. 7
   $$2HI+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4⟶{\mathrm{I}}_2+\mathrm{S}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$$
8. 8
   $$2HBr+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4⟶\mathrm{B}{\mathrm{r}}_2+\mathrm{S}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$$
9. 9I⁻ can also reduce Cu²⁺ to Cu⁺, whereas Br⁻ cannot under the stated comparison:
10. 10
    $$4{\mathrm{I}}^{-}+2C{\mathrm{u}}^{2+}⟶\mathrm{C}{\mathrm{u}}_2{\mathrm{I}}_2+{\mathrm{I}}_2$$
11. 11The reducing-acid order is therefore HF < HCl < HBr < HI, making HI the strongest reductant among the hydrohalic acids.

Step-by-step solution

1. 1In XeO₆⁴⁻, xenon is +8; in XeO₃, xenon is +6. Xenon is therefore reduced.
2. 2Fluoride changes from −1 to 0 in F₂, so fluoride is oxidised.
3. 3
   $$\mathrm{Xe}{\mathrm{O}}_6^{4-}+2{\mathrm{F}}^{-}+6{\mathrm{H}}^{+}⟶\mathrm{Xe}{\mathrm{O}}_3+{\mathrm{F}}_2+3{\mathrm{H}}_2\mathrm{O}$$
4. 4The reaction shows that XeO₆⁴⁻, and hence Na₄XeO₆, can oxidise fluoride to elemental fluorine. Perxenate is therefore a very powerful oxidising agent.

Step-by-step solution

1. 1In (a), P changes from +1 in H₃PO₂ to +5 in H₃PO₄. H₃PO₂ supplies electrons and Ag⁺ is reduced to Ag.
2. 2In (b), P again changes from +1 to +5, but Cu²⁺ is not reduced under these conditions. H₃PO₂ is a stronger reducing agent than benzaldehyde.
3. 3In (c), the aldehyde carbon changes from +1 to +3, and Ag⁺ is reduced to Ag. Benzaldehyde can therefore reduce Ag⁺.
4. 4In (d), benzaldehyde does not reduce Cu²⁺ under the stated conditions. Cu²⁺ is a weaker oxidising agent than Ag⁺ in this comparison.
5. 5The inference is that Ag⁺ is readily reduced by both H₃PO₂ and benzaldehyde, whereas Cu²⁺ is reduced by the stronger reducing agent H₃PO₂ but not by benzaldehyde.

Step-by-step solution

1. 1(a) Basic medium. The half-reactions are I⁻ → I₂ + e⁻ and MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Equating six electrons gives:
2. 2
   $$2Mn{\mathrm{O}}_4^{-}+6{\mathrm{I}}^{-}+4{\mathrm{H}}_2\mathrm{O}⟶2Mn{\mathrm{O}}_2+3{\mathrm{I}}_2+8O{\mathrm{H}}^{-}$$
3. 3(b) Acidic medium. Use MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and SO₂ + 2H₂O → HSO₄⁻ + 3H⁺ + 2e⁻. Equating ten electrons gives:
4. 4
   $$2Mn{\mathrm{O}}_4^{-}+5S{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}^{+}⟶2M{\mathrm{n}}^{2+}+5HS{\mathrm{O}}_4^{-}$$
5. 5(c) Acidic medium. The half-reactions are H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O and Fe²⁺ → Fe³⁺ + e⁻. Thus:
6. 6
   $${\mathrm{H}}_2{\mathrm{O}}_2+2F{\mathrm{e}}^{2+}+2{\mathrm{H}}^{+}⟶2F{\mathrm{e}}^{3+}+2{\mathrm{H}}_2\mathrm{O}$$
7. 7(d) Acidic medium. Use Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O and SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻. Equating six electrons gives:
8. 8
   $$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+3S{\mathrm{O}}_2+2{\mathrm{H}}^{+}⟶2C{\mathrm{r}}^{3+}+3S{\mathrm{O}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}$$

Step-by-step solution

1. 1(a) P is 0 in P₄, −3 in PH₃ and +2 in HPO₂⁻. The basic-medium half-reactions are:
2. 2
   $${\mathrm{P}}_4+12{\mathrm{H}}_2\mathrm{O}+12{\mathrm{e}}^{-}⟶4P{\mathrm{H}}_3+12O{\mathrm{H}}^{-}$$
3. 3
   $$3{\mathrm{P}}_4+36O{\mathrm{H}}^{-}⟶12HP{\mathrm{O}}_2^{-}+12{\mathrm{H}}_2\mathrm{O}+24{\mathrm{e}}^{-}$$
4. 4Adding the appropriately multiplied half-reactions gives:
5. 5
   $$5{\mathrm{P}}_4+12{\mathrm{H}}_2\mathrm{O}+12O{\mathrm{H}}^{-}⟶8P{\mathrm{H}}_3+12HP{\mathrm{O}}_2^{-}$$
6. 6P₄ is both oxidised and reduced, so this is disproportionation; P₄ is the reducing agent for the oxidation branch and the oxidising agent for the reduction branch.
7. 7The oxidation-number check gives 12 electrons gained for four P atoms changing 0 to −3 and 8 electrons lost for four P atoms changing 0 to +2. Equalising 24 electrons requires two P₄ on the reduction branch and three P₄ on the oxidation branch, producing the overall 5P₄ ratio.
8. 8(b) N changes from −2 in N₂H₄ to +2 in NO, while Cl changes from +5 in ClO₃⁻ to −1. Basic half-reactions are:
9. 9
   $${\mathrm{N}}_2{\mathrm{H}}_4+8O{\mathrm{H}}^{-}⟶2NO+6{\mathrm{H}}_2\mathrm{O}+8{\mathrm{e}}^{-}$$
10. 10
    $$\mathrm{Cl}{\mathrm{O}}_3^{-}+3{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}⟶\mathrm{C}{\mathrm{l}}^{-}+6O{\mathrm{H}}^{-}$$
11. 11Equating 24 electrons gives:
12. 12
    $$3{\mathrm{N}}_2{\mathrm{H}}_4+4Cl{\mathrm{O}}_3^{-}⟶6NO+4C{\mathrm{l}}^{-}+6{\mathrm{H}}_2\mathrm{O}$$
13. 13N₂H₄ is the reducing agent and ClO₃⁻ is the oxidising agent.
14. 14The oxidation-number check is N: −2 → +2, so N₂H₄ loses 8 electrons, and Cl: +5 → −1, so each ClO₃⁻ gains 6 electrons. The 24-electron LCM gives 3 N₂H₄ and 4 ClO₃⁻.
15. 15(c) Cl changes from +7 in Cl₂O₇ to +3 in ClO₂⁻, while peroxide oxygen changes from −1 to 0. Basic half-reactions are:
16. 16
    $$\mathrm{C}{\mathrm{l}}_2{\mathrm{O}}_7+3{\mathrm{H}}_2\mathrm{O}+8{\mathrm{e}}^{-}⟶2Cl{\mathrm{O}}_2^{-}+6O{\mathrm{H}}^{-}$$
17. 17
    $${\mathrm{H}}_2{\mathrm{O}}_2+2O{\mathrm{H}}^{-}⟶{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}$$
18. 18Equating eight electrons gives:
19. 19
    $$\mathrm{C}{\mathrm{l}}_2{\mathrm{O}}_7+4{\mathrm{H}}_2{\mathrm{O}}_2+2O{\mathrm{H}}^{-}⟶2Cl{\mathrm{O}}_2^{-}+4{\mathrm{O}}_2+5{\mathrm{H}}_2\mathrm{O}$$
20. 20Cl₂O₇ is the oxidising agent and H₂O₂ is the reducing agent.
21. 21The oxidation-number check is Cl: +7 → +3, so each Cl₂O₇ gains 8 electrons, and peroxide O: −1 → 0, so each H₂O₂ loses 2 electrons. This gives one Cl₂O₇ for every four H₂O₂ before balancing O, H and charge.

Step-by-step solution

1. 1The reaction takes place in alkaline medium and is a disproportionation reaction of cyanogen, (CN)₂.
2. 2The carbon oxidation state is +3 in cyanogen, decreases to +2 in CN⁻ and increases to +4 in CNO⁻.
3. 3Thus cyanogen is oxidised and reduced simultaneously, showing that the C–N unit can undergo both reduction and oxidation.
4. 4The reaction also demonstrates the reducing and oxidising character of cyanogen in alkaline solution and the formation of cyanide and cyanate ions.
5. 5
   $$(\mathrm{CN}{)}_2+2O{\mathrm{H}}^{-}⟶\mathrm{C}{\mathrm{N}}^{-}+\mathrm{CN}{\mathrm{O}}^{-}+{\mathrm{H}}_2\mathrm{O}$$

Step-by-step solution

1. 1Mn is +3 initially. It is reduced to Mn²⁺ and oxidised to Mn in MnO₂ at +4.
2. 2
   $$\mathrm{M}{\mathrm{n}}^{3+}+{\mathrm{e}}^{-}⟶\mathrm{M}{\mathrm{n}}^{2+}$$
3. 3
   $$\mathrm{M}{\mathrm{n}}^{3+}+2{\mathrm{H}}_2\mathrm{O}⟶\mathrm{Mn}{\mathrm{O}}_2+4{\mathrm{H}}^{+}+{\mathrm{e}}^{-}$$
4. 4Add the half-reactions; two Mn³⁺ ions undergo disproportionation:
5. 5
   $$2M{\mathrm{n}}^{3+}+2{\mathrm{H}}_2\mathrm{O}⟶\mathrm{M}{\mathrm{n}}^{2+}+\mathrm{Mn}{\mathrm{O}}_2+4{\mathrm{H}}^{+}$$
6. 6Because H⁺ is a product, a high H⁺ concentration suppresses the forward reaction; dilution or removal of H⁺ favours disproportionation.

Step-by-step solution

1. 1(a) F is −1 in all ordinary compounds and is the element with only negative oxidation state.
2. 2(b) Cs is an alkali metal and has oxidation state +1 in all its compounds.
3. 3(c) I shows −1 in iodides and positive states such as +1, +5 and +7 in oxycompounds.
4. 4(d) Ne is a noble gas with oxidation state 0 and does not show either positive or negative oxidation state under ordinary conditions.
5. 5The relevant oxidation-number assignments are F = −1, Cs = +1 and Ne = 0.

Step-by-step solution

1. 1Cl₂ is reduced from 0 to −1, while S in SO₂ is oxidised from +4 to +6 in sulphate.
2. 2The ionic redox change is:
3. 3
   $$\mathrm{C}{\mathrm{l}}_2+\mathrm{S}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}⟶2C{\mathrm{l}}^{-}+\mathrm{S}{\mathrm{O}}_4^{2-}+4{\mathrm{H}}^{+}$$
4. 4A molecular form is:
5. 5
   $$\mathrm{C}{\mathrm{l}}_2+\mathrm{S}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}⟶2HCl+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$$
6. 6Chlorine is the oxidising agent and SO₂ is the reducing agent.

Step-by-step solution

1. 1A disproportionation reaction contains the same element in an intermediate oxidation state in both the oxidised and reduced products.
2. 2(a) P, Cl and S are suitable non-metals. For example, P(0) can form P(−3) and P(+1), Cl(0) can form Cl(−1) and Cl(+1), and S(0) can form S(−2) and S(+4).
3. 3(b) Mn, Cu and Ga are suitable metals. Mn(III) can form Mn(II) and Mn(IV); Cu(I) can form Cu(0) and Cu(II); Ga can show Ga(0), Ga(I) and Ga(III) behaviour.
4. 4The selected element must have access to at least three relevant oxidation states, with one state between the products of oxidation and reduction.

Step-by-step solution

1. 1Moles of ammonia:
2. 2
   $$n(\mathrm{N}{\mathrm{H}}_3)=\frac{10.00\mathrm{g}}{17.00\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}}=0.5882\mathrm{mol}$$
3. 3The O₂ required for this amount is:
4. 4
   $$n({\mathrm{O}}_2{)}_{\text{required}}=\frac{5}{4}(0.5882)=0.7353\mathrm{mol}$$
5. 5
   $$m({\mathrm{O}}_2{)}_{\text{required}}=0.7353(32.00)=23.53\mathrm{g}$$
6. 6Only 20.00 g O₂ is available, so O₂ is the limiting reagent.
7. 7Moles of O₂ available:
8. 8
   $$n({\mathrm{O}}_2)=\frac{20.00\mathrm{g}}{32.00\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}}=0.6250\mathrm{mol}$$
9. 9From 5 mol O₂, 4 mol NO are formed:
10. 10
    $$n(\mathrm{NO})=\frac{4}{5}(0.6250)=0.5000\mathrm{mol}$$
11. 11Mass of NO:
12. 12
    $$m(\mathrm{NO})=0.5000(30.00)=15.00\mathrm{g}$$

Step-by-step solution

1. 1For a spontaneous redox reaction, E°cell = E°cathode − E°anode must be positive. Do not multiply E° values by coefficients in the balanced equation.
2. 2(a) Fe³⁺ is reduced and I⁻ is oxidised: E°cell = 0.77 − 0.54 = +0.23 V. The reaction is feasible.
3. 3(b) Ag⁺ is reduced and Cu is oxidised: E°cell = 0.80 − 0.34 = +0.46 V. The reaction is feasible.
4. 4(c) Fe³⁺ is reduced and Cu is oxidised: E°cell = 0.77 − 0.34 = +0.43 V. The reaction is feasible.
5. 5(d) Fe³⁺ is reduced and Ag is oxidised: E°cell = 0.77 − 0.80 = −0.03 V. The reaction is not feasible in the stated direction.
6. 6(e) Br₂ is reduced and Fe²⁺ is oxidised: E°cell = 1.09 − 0.77 = +0.32 V. The reaction is feasible.

Step-by-step solution

1. 1At the cathode, reduction occurs; at the anode, oxidation occurs. The electrode material can supply a species more readily than water or the dissolved ions.
2. 2(i) With Ag electrodes, Ag⁺ is reduced at the cathode and Ag metal is oxidised at the anode. Silver is transferred from anode to cathode and the electrolyte concentration remains essentially unchanged.
3. 3(ii) With Pt electrodes, Ag⁺ is reduced to Ag at the cathode, while water is oxidised to O₂ at the anode:
4. 4
   $$4AgN{\mathrm{O}}_3+2{\mathrm{H}}_2\mathrm{O}⟶4Ag+{\mathrm{O}}_2+4HN{\mathrm{O}}_3$$
5. 5(iii) In dilute H₂SO₄, H⁺ is reduced to H₂ at the cathode and water is oxidised to O₂ at the anode:
6. 6
   $$2{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{H}}_2+{\mathrm{O}}_2$$
7. 7(iv) Cu²⁺ is reduced to Cu at the cathode and Cl⁻ is oxidised to Cl₂ at the anode:
8. 8
   $$\mathrm{CuC}{\mathrm{l}}_2⟶\mathrm{Cu}+\mathrm{C}{\mathrm{l}}_2$$

Step-by-step solution

1. 1A metal displaces another metal from its salt solution when it has a greater tendency to be oxidised, or equivalently when the displaced metal has a more negative standard reduction potential.
2. 2The relevant reduction potentials decrease in the order Mg²⁺/Mg, Al³⁺/Al, Zn²⁺/Zn, Fe²⁺/Fe and Cu²⁺/Cu.
3. 3Therefore the displacement or reducing-power order is:
4. 4
   $$\mathrm{Mg}>\mathrm{Al}>\mathrm{Zn}>\mathrm{Fe}>\mathrm{Cu}$$
5. 5Each metal to the left can displace a metal to its right from a solution of the latter's salt.

Step-by-step solution

1. 1Reducing power is the tendency of a metal to lose electrons and is inversely related to the magnitude of its standard reduction potential.
2. 2The more negative the reduction potential, the stronger the metal is as a reducing agent. The given order from least to most reducing is therefore:
3. 3
   $$\mathrm{Ag}<\mathrm{Hg}<\mathrm{Cr}<\mathrm{Mg}<\mathrm{K}$$
4. 4For example, K has E° = −2.93 V and is more easily oxidised than Mg with E° = −2.37 V, while Ag with E° = +0.80 V is the least reducing in the set.

Step-by-step solution

1. 1Zinc is oxidised at the anode and silver ions are reduced at the cathode:
2. 2
   $$\mathrm{Zn}(\mathrm{s})⟶\mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}$$
3. 3
   $$2A{\mathrm{g}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}⟶2Ag(\mathrm{s})$$
4. 4The zinc electrode is therefore the negative electrode (anode), and the silver electrode is the positive electrode (cathode).
5. 5Electrons travel through the external circuit from the Zn electrode to the Ag electrode. In the salt bridge, anions migrate toward the Zn anode and cations migrate toward the Ag cathode to maintain electrical neutrality.
6. 6The cell can be represented in words as Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s). Using E°Ag⁺/Ag = 0.80 V and E°Zn²⁺/Zn = −0.76 V:
7. 7
   $$E_{\mathrm{cell}}^{\circ }=E_{\mathrm{cathode}}^{\circ }-E_{\mathrm{anode}}^{\circ }=0.80-(-0.76)=+1.56\mathrm{V}$$