Balancing Redox Equations

AS · 13 min

Redox equations are harder to balance by inspection than ordinary equations, because the electrons lost must exactly equal the electrons gained, and many reactions in acid involve water and hydrogen ions that do not appear in a word equation. This note gives two reliable methods: building and combining half-equations, and using changes in oxidation number. It then applies them to the redox titrations (manganate(VII), dichromate(VI) and iodine–thiosulfate) that appear in Paper 2 calculations and Paper 3 practicals.

Half-equations

A half-equation shows either the oxidation or the reduction part of a redox reaction, with electrons written explicitly. Electrons appear on the left in a reduction (they are gained) and on the right in an oxidation (they are lost).

FeX3++eX−→FeX2+(reduction)\ce{Fe^3+ + e- -> Fe^2+} \qquad \text{(reduction)} 2 IX−→IX2+2 eX−(oxidation)\ce{2I- -> I2 + 2e-} \qquad \text{(oxidation)}

A half-equation must balance in atoms and in charge. The electrons are what balance the charge.

Half-equations for oxyanions in acid

Many oxidising agents are oxyanions used in acidic solution, such as manganate(VII) and dichromate(VI). Their half-equations need water and hydrogen ions.

Method

Building a half-equation in acidic solution

  1. Write the species that changes and what it becomes: MnOX4X−→MnX2+\ce{MnO4- -> Mn^2+}.
  2. Balance the atoms of the element that changes oxidation number.
  3. Balance oxygen by adding HX2O\ce{H2O} to the side short of oxygen.
  4. Balance hydrogen by adding HX+\ce{H+} to the side short of hydrogen.
  5. Balance charge by adding electrons (eX−\ce{e-}) to the more positive side.
  6. Check atoms and charge.

For manganate(VII):

  1. MnOX4X−→MnX2+\ce{MnO4- -> Mn^2+}
  2. Mn is balanced.
  3. Four O on the left: add 4 HX2O\ce{4H2O} on the right: MnOX4X−→MnX2++4 HX2O\ce{MnO4- -> Mn^2+ + 4H2O}.
  4. Eight H on the right: add 8 HX+\ce{8H+} on the left: MnOX4X−+8 HX+→MnX2++4 HX2O\ce{MnO4- + 8H+ -> Mn^2+ + 4H2O}.
  5. Charge on the left: −1+8=+7-1 + 8 = +7; on the right: +2+2. Add 5eX−5\ce{e-} to the left.
MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}

Check with oxidation numbers: Mn goes from +7+7 to +2+2, a gain of 5 electrons. It matches.

Key result

Half-equations to know

specieshalf-equation
manganate(VII), acidMnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}
dichromate(VI), acidCrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}
iron(II)FeX2+→FeX3++eX−\ce{Fe^2+ -> Fe^3+ + e-}
iodide2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-}
thiosulfate2 SX2OX3X2−→SX4OX6X2−+2 eX−\ce{2S2O3^2- -> S4O6^2- + 2e-}
ethanedioateCX2OX4X2−→2 COX2+2 eX−\ce{C2O4^2- -> 2CO2 + 2e-}
hydrogen peroxide as oxidising agentHX2OX2+2 HX++2 eX−→2 HX2O\ce{H2O2 + 2H+ + 2e- -> 2H2O}
hydrogen peroxide as reducing agentHX2OX2→OX2+2 HX++2 eX−\ce{H2O2 -> O2 + 2H+ + 2e-}
sulfur dioxideSOX2+2 HX2O→SOX4X2−+4 HX++2 eX−\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}
nitrate to NONOX3X−+4 HX++3 eX−→NO+2 HX2O\ce{NO3- + 4H+ + 3e- -> NO + 2H2O}

You do not need to memorise all of these: you should be able to build any of them with the method above.

Combining half-equations

Method

Writing the full ionic equation from two half-equations

  1. Write the reduction and oxidation half-equations.
  2. Multiply one or both so that the number of electrons is the same in each.
  3. Add them together. The electrons cancel.
  4. Cancel any HX+\ce{H+}, HX2O\ce{H2O} or other species that appear on both sides.
  5. Check atoms and charge.

Manganate(VII) with iron(II). Reduction gains 5 electrons; oxidation loses 1. Multiply the iron half-equation by 5:

MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} 5 FeX2+→5 FeX3++5 eX−\ce{5Fe^2+ -> 5Fe^3+ + 5e-}

Add:

MnOX4X−(aq)+8 HX+(aq)+5 FeX2+(aq)→MnX2+(aq)+4 HX2O(l)+5 FeX3+(aq)\ce{MnO4-(aq) + 8H+(aq) + 5Fe^2+(aq) -> Mn^2+(aq) + 4H2O(l) + 5Fe^3+(aq)}

Charge check: left −1+8+10=+17-1 + 8 + 10 = +17; right +2+15=+17+2 + 15 = +17.

Dichromate(VI) with iron(II):

CrX2OX7X2−(aq)+14 HX+(aq)+6 FeX2+(aq)→2 CrX3+(aq)+7 HX2O(l)+6 FeX3+(aq)\ce{Cr2O7^2-(aq) + 14H+(aq) + 6Fe^2+(aq) -> 2Cr^3+(aq) + 7H2O(l) + 6Fe^3+(aq)}

Alkaline conditions

Tip

If a reaction happens in alkaline solution, balance it in acid first, then add the same number of OHX−\ce{OH-} ions to both sides as there are HX+\ce{H+} ions; combine HX++OHX−\ce{H+ + OH-} into HX2O\ce{H2O} and cancel water from both sides. At AS level, most redox equations you balance are in acid or need no HX+\ce{H+} at all.

The oxidation number method

When you are given the reactants and products of a full equation (not ionic), it is often quicker to balance using oxidation numbers. The principle: the total increase in oxidation number must equal the total decrease.

Method

Balancing with changes in oxidation number

  1. Write the unbalanced equation and identify the elements whose oxidation numbers change.
  2. Work out the change per atom for each, and per formula unit.
  3. Choose multipliers so that the total increase equals the total decrease.
  4. Put those coefficients in front of the species that change.
  5. Balance the remaining atoms (spectator ions, then O with HX2O\ce{H2O} and H with HX+\ce{H+} if ionic).
  6. Check atoms and charge.

Copper with dilute nitric acid. Unbalanced: Cu+HNOX3→Cu(NOX3)X2+NO+HX2O\ce{Cu + HNO3 -> Cu(NO3)2 + NO + H2O}.

  • Cu: 0→+20 \to +2, increase of 2.
  • N (the nitrogen that ends up in NO): +5→+2+5 \to +2, decrease of 3.

The lowest common multiple of 2 and 3 is 6, so 3 Cu (increase 6) and 2 NO (decrease 6):

3 Cu+?HNOX3→3 Cu(NOX3)X2+2 NO+?HX2O\ce{3Cu + ?HNO3 -> 3Cu(NO3)2 + 2NO + ?H2O}

Nitrogen on the right: 3×2=63 \times 2 = 6 in nitrate plus 2 in NO =8= 8, so 8 HNOX3\ce{8HNO3}. Hydrogen: 8 on the left, so 4 HX2O\ce{4H2O}. Oxygen check: left 2424; right 18+2+4=2418 + 2 + 4 = 24.

3 Cu(s)+8 HNOX3(aq)→3 Cu(NOX3)X2(aq)+2 NO(g)+4 HX2O(l)\ce{3Cu(s) + 8HNO3(aq) -> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)}

The method works equally well for disproportionation. For hot alkali, ClX2+NaOH→NaCl+NaClOX3+HX2O\ce{Cl2 + NaOH -> NaCl + NaClO3 + H2O}: Cl to NaCl\ce{NaCl} decreases by 1; Cl to NaClOX3\ce{NaClO3} increases by 5. So 5 NaCl\ce{NaCl} for each NaClOX3\ce{NaClO3}: six Cl atoms, 3 ClX2\ce{3Cl2}; then 6 NaOH\ce{6NaOH} and 3 HX2O\ce{3H2O}:

3 ClX2+6 NaOH→5 NaCl+NaClOX3+3 HX2O\ce{3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O}

Redox titrations

Balanced redox equations give the mole ratios for titration calculations, exactly as in acid–base titrations.

Potassium manganate(VII) titrations. Acidified KMnOX4\ce{KMnO4} is purple; MnX2+\ce{Mn^2+} is very pale pink, effectively colourless. The manganate(VII) is in the burette. As it is added to the reducing agent (for example FeX2+\ce{Fe^2+}) it is decolourised; the end point is the first permanent pale pink colour, when one drop of excess MnOX4X−\ce{MnO4-} remains. No indicator is needed: the titration is self-indicating. The flask is acidified with dilute sulfuric acid (not hydrochloric acid, which manganate(VII) would oxidise to chlorine).

Iodine–thiosulfate titrations. An oxidising agent is added to excess potassium iodide, releasing iodine. The iodine is titrated with sodium thiosulfate:

IX2(aq)+2 SX2OX3X2−(aq)→2 IX−(aq)+SX4OX6X2−(aq)\ce{I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq)}

The brown iodine fades to pale yellow; starch is then added, giving a blue-black colour, and the end point is when the blue-black colour disappears. Starch is added near the end, not at the start, because a large amount of iodine bound to starch is released only slowly.

Worked examples

Building a half-equation

Write the half-equation for the reduction of dichromate(VI) ions to chromium(III) ions in acidic solution.

Solution
  1. CrX2OX7X2−→CrX3+\ce{Cr2O7^2- -> Cr^3+}
  2. Balance Cr: CrX2OX7X2−→2 CrX3+\ce{Cr2O7^2- -> 2Cr^3+}
  3. Balance O with water: CrX2OX7X2−→2 CrX3++7 HX2O\ce{Cr2O7^2- -> 2Cr^3+ + 7H2O}
  4. Balance H with HX+\ce{H+}: CrX2OX7X2−+14 HX+→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ -> 2Cr^3+ + 7H2O}
  5. Charge: left −2+14=+12-2 + 14 = +12; right +6+6. Add 6eX−6\ce{e-} to the left.
CrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}

Check: each Cr goes from +6+6 to +3+3, gaining 3 electrons; two Cr gain 6.

Combining half-equations

Acidified potassium manganate(VII) oxidises ethanedioate ions, CX2OX4X2−\ce{C2O4^2-}, to carbon dioxide. Write the ionic equation.

Solution

Reduction: MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} (5 electrons).

Oxidation: CX2OX4X2−→2 COX2+2 eX−\ce{C2O4^2- -> 2CO2 + 2e-} (2 electrons).

LCM of 5 and 2 is 10: multiply the reduction by 2 and the oxidation by 5.

2 MnOX4X−+16 HX++5 CX2OX4X2−→2 MnX2++8 HX2O+10 COX2\ce{2MnO4- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2}

Charge check: left −2+16−10=+4-2 + 16 - 10 = +4; right +4+4.

Oxidation number method

Balance: CrX2OX7X2−+SOX2+HX+→CrX3++SOX4X2−+HX2O\ce{Cr2O7^2- + SO2 + H+ -> Cr^3+ + SO4^2- + H2O}.

Solution

Cr: +6→+3+6 \to +3, decrease of 3 per Cr, 6 per CrX2OX7X2−\ce{Cr2O7^2-}.

S: +4→+6+4 \to +6, increase of 2 per SOX2\ce{SO2}.

To balance 6 with 2: 3 SOX2\ce{SO2} per CrX2OX7X2−\ce{Cr2O7^2-}:

CrX2OX7X2−+3 SOX2+?HX+→2 CrX3++3 SOX4X2−+?HX2O\ce{Cr2O7^2- + 3SO2 + ?H+ -> 2Cr^3+ + 3SO4^2- + ?H2O}

Oxygen: left 7+6=137 + 6 = 13; right 1212 + water. So 1 HX2O\ce{1H2O}. Hydrogen: right 2, so 2 HX+\ce{2H+}.

CrX2OX7X2−+3 SOX2+2 HX+→2 CrX3++3 SOX4X2−+HX2O\ce{Cr2O7^2- + 3SO2 + 2H+ -> 2Cr^3+ + 3SO4^2- + H2O}

Charge: left −2+2=0-2 + 2 = 0; right +6−6=0+6 - 6 = 0. Balanced.

Manganate(VII) titration

An iron tablet of mass 0.800 g0.800\ \text{g} was dissolved in dilute sulfuric acid. The solution required 22.50 cm322.50\ \text{cm}^3 of 0.0200 mol dm−30.0200\ \text{mol dm}^{-3} potassium manganate(VII) for complete reaction. Calculate the percentage by mass of iron in the tablet.

Solution

n(MnOX4X−)=0.02250×0.0200=4.50×10−4 moln(\ce{MnO4-}) = 0.02250 \times 0.0200 = 4.50 \times 10^{-4}\ \text{mol}.

Ratio MnOX4X−:FeX2+=1:5\ce{MnO4-} : \ce{Fe^2+} = 1 : 5, so n(FeX2+)=5×4.50×10−4=2.25×10−3 moln(\ce{Fe^2+}) = 5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3}\ \text{mol}.

Mass of iron =2.25×10−3×55.8=0.1256 g= 2.25 \times 10^{-3} \times 55.8 = 0.1256\ \text{g}.

% Fe=0.12560.800×100=15.7%\%\ \text{Fe} = \frac{0.1256}{0.800} \times 100 = 15.7\%
Exam-hard: concentration of hydrogen peroxide

10.0 cm310.0\ \text{cm}^3 of a hydrogen peroxide solution was diluted to 250 cm3250\ \text{cm}^3 in a volumetric flask. 25.0 cm325.0\ \text{cm}^3 of the diluted solution, acidified with dilute sulfuric acid, needed 24.00 cm324.00\ \text{cm}^3 of 0.0200 mol dm−30.0200\ \text{mol dm}^{-3} potassium manganate(VII) to reach the end point.

(a) Construct the overall ionic equation. (b) Calculate the concentration of the original hydrogen peroxide. (c) State the colour change at the end point and explain why no indicator is needed.

Solution

(a) Reduction: MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} (×2\times 2). Oxidation: HX2OX2→OX2+2 HX++2 eX−\ce{H2O2 -> O2 + 2H+ + 2e-} (×5\times 5).

2 MnOX4X−+16 HX++5 HX2OX2→2 MnX2++8 HX2O+5 OX2+10 HX+\ce{2MnO4- + 16H+ + 5H2O2 -> 2Mn^2+ + 8H2O + 5O2 + 10H+}

Cancel 10HX+10\ce{H+}:

2 MnOX4X−(aq)+6 HX+(aq)+5 HX2OX2(aq)→2 MnX2+(aq)+5 OX2(g)+8 HX2O(l)\ce{2MnO4-(aq) + 6H+(aq) + 5H2O2(aq) -> 2Mn^2+(aq) + 5O2(g) + 8H2O(l)}

(b) n(MnOX4X−)=0.02400×0.0200=4.80×10−4 moln(\ce{MnO4-}) = 0.02400 \times 0.0200 = 4.80 \times 10^{-4}\ \text{mol}.

n(HX2OX2)n(\ce{H2O2}) in 25.0 cm325.0\ \text{cm}^3 =52×4.80×10−4=1.20×10−3 mol= \tfrac{5}{2} \times 4.80 \times 10^{-4} = 1.20 \times 10^{-3}\ \text{mol}.

In 250 cm3250\ \text{cm}^3 (which contains all the original 10.0 cm310.0\ \text{cm}^3): 1.20×10−2 mol1.20 \times 10^{-2}\ \text{mol}.

Concentration of the original =1.20×10−20.0100=1.20 mol dm−3= \dfrac{1.20 \times 10^{-2}}{0.0100} = 1.20\ \text{mol dm}^{-3}.

(c) Colourless to the first permanent pale pink. Manganate(VII) is intensely purple and its reduction product, MnX2+\ce{Mn^2+}, is almost colourless, so the first drop of excess MnOX4X−\ce{MnO4-} colours the solution: it acts as its own indicator.

Redox titrations

Manganate(VII) titration (for example, FeX2+\ce{Fe^2+} in iron(II) ammonium sulfate or an iron tablet)

  1. Pipette 25.0 cm325.0\ \text{cm}^3 of the FeX2+\ce{Fe^2+} solution into a conical flask and add about 10 cm310\ \text{cm}^3 of dilute sulfuric acid (excess).
  2. Fill the burette with potassium manganate(VII). Because the solution is so intensely coloured, read the top of the meniscus.
  3. Titrate until the first permanent pale pink colour; repeat until two concordant titres (within 0.10 cm30.10\ \text{cm}^3) are obtained.

Iodine–thiosulfate titration (for example, finding the concentration of an oxidising agent such as IOX3X−\ce{IO3-}, CuX2+\ce{Cu^2+} or ClOX−\ce{ClO-})

  1. Pipette the oxidising agent into a flask; add excess potassium iodide (and acid if required). Iodine is released.
  2. Titrate with sodium thiosulfate until the brown colour becomes pale yellow.
  3. Add a few drops of starch: the solution turns blue-black. Continue dropwise until the blue-black colour just disappears.

Points examiners ask about: why the acid is sulfuric not hydrochloric; why the manganate(VII) meniscus is read at the top; why starch is added near the end point; why potassium iodide is added in excess (so all the oxidising agent reacts and the iodine stays in solution as IX3X−\ce{I3-}); the percentage uncertainty in a titre (burette ±0.05 cm3\pm 0.05\ \text{cm}^3 per reading, so ±0.10 cm3\pm 0.10\ \text{cm}^3 per titre).

Watch out
  • Electrons on the wrong side. Reduction: electrons on the left. Oxidation: electrons on the right.
  • Not cancelling electrons. The final ionic equation must contain no electrons. Multiply until they are equal, then cancel.
  • Forgetting to cancel HX+\ce{H+} and HX2O\ce{H2O} that appear on both sides after adding.
  • Balancing atoms but not charge. Always do a final charge check; it catches most errors.
  • Using HCl to acidify manganate(VII). Manganate(VII) oxidises chloride ions to chlorine, using up extra MnOX4X−\ce{MnO4-} and giving a titre that is too large.
Exam tip
  • "Construct an ionic equation from these half-equations" is worth 1 to 2 marks; the multipliers and the cancelled final equation must both be right.
  • In titration calculations, write the mole ratio from the equation explicitly ("MnOX4X−:FeX2+=1:5\ce{MnO4-} : \ce{Fe^2+} = 1 : 5") before using it.
  • Show scaling-up steps clearly (for example from 25.0 cm325.0\ \text{cm}^3 to 250 cm3250\ \text{cm}^3, then to the original sample).
  • Give the colour change at an end point in the right order (from … to …): manganate(VII): colourless (or pale yellow-green if FeX3+\ce{Fe^3+} is present) to pale pink; thiosulfate with starch: blue-black to colourless.
Summary
  • Half-equations show electrons: on the left for reduction, on the right for oxidation; balance atoms and charge.
  • In acid: balance the changing element, then O with HX2O\ce{H2O}, H with HX+\ce{H+}, and charge with eX−\ce{e-}.
  • Combine half-equations by equalising electrons, adding, and cancelling.
  • Oxidation number method: total increase = total decrease; then balance the rest.
  • MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} (1 MnOX4X−\ce{MnO4-} : 5 FeX2+\ce{Fe^2+}); CrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O} (1 : 6 FeX2+\ce{Fe^2+}); IX2+2 SX2OX3X2−→2 IX−+SX4OX6X2−\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}.
  • Manganate(VII) titrations are self-indicating (to pale pink) and use dilute sulfuric acid; iodine–thiosulfate uses starch near the end point (blue-black to colourless).

Practice

Question
  1. Write half-equations for: (a) ClX2\ce{Cl2} being reduced to ClX−\ce{Cl-}; (b) FeX2+\ce{Fe^2+} being oxidised; (c) NOX3X−\ce{NO3-} being reduced to NOX2\ce{NO2} in acid.
  2. Construct the ionic equation for the oxidation of iodide ions by hydrogen peroxide in acid, using HX2OX2+2 HX++2 eX−→2 HX2O\ce{H2O2 + 2H+ + 2e- -> 2H2O}.
  3. Combine the half-equations for iron(III) and iodide to write the equation for the reaction of FeX3+\ce{Fe^3+} with IX−\ce{I-}.
  4. Use oxidation numbers to balance KMnOX4+HCl→KCl+MnClX2+ClX2+HX2O\ce{KMnO4 + HCl -> KCl + MnCl2 + Cl2 + H2O}.
  5. Balance IOX3X−+IX−+HX+→IX2+HX2O\ce{IO3- + I- + H+ -> I2 + H2O} and state what type of redox reaction it is from the point of view of iodine.
  6. 25.0 cm325.0\ \text{cm}^3 of an iron(II) solution needed 20.00 cm320.00\ \text{cm}^3 of 0.0150 mol dm−30.0150\ \text{mol dm}^{-3} potassium dichromate(VI). Calculate the concentration of FeX2+\ce{Fe^2+}.
  7. 0.150 g0.150\ \text{g} of sodium ethanedioate, NaX2CX2OX4\ce{Na2C2O4} (Mr=134.0M_r = 134.0), was dissolved and acidified; it needed 22.40 cm322.40\ \text{cm}^3 of potassium manganate(VII) solution. Calculate the concentration of the manganate(VII).
  8. 25.0 cm325.0\ \text{cm}^3 of 0.0100 mol dm−30.0100\ \text{mol dm}^{-3} potassium iodate(V) was added to excess acidified potassium iodide. The iodine released needed 18.75 cm318.75\ \text{cm}^3 of sodium thiosulfate. Using your equation from question 5, calculate the concentration of the thiosulfate.
  9. Iron(II) ions are oxidised by oxygen in acidic solution. Construct the overall equation from the half-equations OX2+4 HX++4 eX−→2 HX2O\ce{O2 + 4H+ + 4e- -> 2H2O} and FeX2+→FeX3++eX−\ce{Fe^2+ -> Fe^3+ + e-}, and explain why a solution of iron(II) sulfate used for titration is made up freshly.
  10. When concentrated sulfuric acid is added to solid potassium bromide, some of the hydrogen bromide formed is oxidised: HBr+HX2SOX4→BrX2+SOX2+HX2O\ce{HBr + H2SO4 -> Br2 + SO2 + H2O}. Balance this equation using oxidation numbers, and explain why a student who titrates an iron(II) solution with manganate(VII) after acidifying with hydrochloric acid obtains a titre that is too large.
Answers
  1. (a) ClX2+2 eX−→2 ClX−\ce{Cl2 + 2e- -> 2Cl-}. (b) FeX2+→FeX3++eX−\ce{Fe^2+ -> Fe^3+ + e-}. (c) NOX3X−+2 HX++eX−→NOX2+HX2O\ce{NO3- + 2H+ + e- -> NO2 + H2O}.
  2. Oxidation: 2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-}. Both have 2 electrons: HX2OX2+2 HX++2 IX−→IX2+2 HX2O\ce{H2O2 + 2H+ + 2I- -> I2 + 2H2O}.
  3. FeX3++eX−→FeX2+\ce{Fe^3+ + e- -> Fe^2+} (×2\times 2) and 2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-}: 2 FeX3++2 IX−→2 FeX2++IX2\ce{2Fe^3+ + 2I- -> 2Fe^2+ + I2}.
  4. Mn: +7→+2+7 \to +2, decrease 5. Cl (to ClX2\ce{Cl2}): −1→0-1 \to 0, increase 1 per atom, 2 per ClX2\ce{Cl2}. LCM 10: 2KMnOX42\ce{KMnO4} and 5ClX25\ce{Cl2}. Then K and Mn: 2KCl2\ce{KCl}, 2MnClX22\ce{MnCl2}. Total Cl on right =2+4+10=16= 2 + 4 + 10 = 16: 16HCl16\ce{HCl}. H: 1616, so 8HX2O8\ce{H2O}; O check 8=88 = 8. 2 KMnOX4+16 HCl→2 KCl+2 MnClX2+5 ClX2+8 HX2O\ce{2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 5Cl2 + 8H2O}.
  5. I in IOX3X−\ce{IO3-}: +5→0+5 \to 0 (decrease 5); I in IX−\ce{I-}: −1→0-1 \to 0 (increase 1). So 1 IOX3X−:5 IX−1\ \ce{IO3-} : 5\ \ce{I-}, giving 3IX23\ce{I2}. O: 3, so 3HX2O3\ce{H2O}; H: 6, so 6HX+6\ce{H+}. IOX3X−+5 IX−+6 HX+→3 IX2+3 HX2O\ce{IO3- + 5I- + 6H+ -> 3I2 + 3H2O}; charge: −1−5+6=0-1 - 5 + 6 = 0. Iodine in two different species (one oxidised, one reduced) forms the same product: the reverse of disproportionation (sometimes called comproportionation).
  6. n(CrX2OX7X2−)=0.02000×0.0150=3.00×10−4 moln(\ce{Cr2O7^2-}) = 0.02000 \times 0.0150 = 3.00 \times 10^{-4}\ \text{mol}. Ratio 1:61 : 6: n(FeX2+)=1.80×10−3 moln(\ce{Fe^2+}) = 1.80 \times 10^{-3}\ \text{mol}. Concentration =1.80×10−30.0250=0.0720 mol dm−3= \dfrac{1.80 \times 10^{-3}}{0.0250} = 0.0720\ \text{mol dm}^{-3}.
  7. n(CX2OX4X2−)=0.150134.0=1.119×10−3 moln(\ce{C2O4^2-}) = \dfrac{0.150}{134.0} = 1.119 \times 10^{-3}\ \text{mol}. Ratio MnOX4X−:CX2OX4X2−=2:5\ce{MnO4-} : \ce{C2O4^2-} = 2 : 5: n(MnOX4X−)=4.478×10−4 moln(\ce{MnO4-}) = 4.478 \times 10^{-4}\ \text{mol}. Concentration =4.478×10−40.02240=0.0200 mol dm−3= \dfrac{4.478 \times 10^{-4}}{0.02240} = 0.0200\ \text{mol dm}^{-3}.
  8. n(IOX3X−)=0.0250×0.0100=2.50×10−4 moln(\ce{IO3-}) = 0.0250 \times 0.0100 = 2.50 \times 10^{-4}\ \text{mol}. n(IX2)=3×2.50×10−4=7.50×10−4 moln(\ce{I2}) = 3 \times 2.50 \times 10^{-4} = 7.50 \times 10^{-4}\ \text{mol}. n(SX2OX3X2−)=2×7.50×10−4=1.50×10−3 moln(\ce{S2O3^2-}) = 2 \times 7.50 \times 10^{-4} = 1.50 \times 10^{-3}\ \text{mol}. Concentration =1.50×10−30.01875=0.0800 mol dm−3= \dfrac{1.50 \times 10^{-3}}{0.01875} = 0.0800\ \text{mol dm}^{-3}.
  9. Multiply the iron half-equation by 4: 4 FeX2++OX2+4 HX+→4 FeX3++2 HX2O\ce{4Fe^2+ + O2 + 4H+ -> 4Fe^3+ + 2H2O}. Iron(II) is slowly oxidised by dissolved oxygen from the air; an old solution contains less FeX2+\ce{Fe^2+} than its label suggests, so it must be made up freshly (and kept acidified) for accurate results.
  10. Br: −1→0-1 \to 0, increase 1 per atom, 2 per BrX2\ce{Br2}. S: +6→+4+6 \to +4, decrease 2. So 2HBr2\ce{HBr} per HX2SOX4\ce{H2SO4}: 2 HBr+HX2SOX4→BrX2+SOX2+2 HX2O\ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}. With HCl, the manganate(VII) also oxidises chloride ions to chlorine (2 MnOX4X−+16 HX++10 ClX−→2 MnX2++5 ClX2+8 HX2O\ce{2MnO4- + 16H+ + 10Cl- -> 2Mn^2+ + 5Cl2 + 8H2O}), so extra manganate(VII) is used up in addition to that needed for the iron(II): the titre is too large and the calculated FeX2+\ce{Fe^2+} concentration too high.

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