MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (aq)
purple colourless
A health supplement tablet contain iron(II)sulfate was analysed by titration. A tablet weighing 2.25 g was dissolved in dilute sulfuric acid and titrated against 0.100 mol dm-3 KMnO4 .The titration required 26.50 cm3 for complete reaction. Calculate the percentage by mass of iron in the table.
Answer
Step 1: Write the balanced equation for the reaction
oxidation: Fe2+ (aq) → Fe3+ (aq) + e-
reduction: MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (l)
overall: MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) → Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)
Step 2: Determine the amount of MnO4- used in the titration
moles of MnO4- = 0.0265 dm3 x 0.100 mol dm-3 = 0.00265 mol
Step 3: Determine the amount of iron in the reaction
From the equation for the reaction we know the reacting ratio MnO4- : Fe2+ = 1: 5
∴ moles of Fe2+ = 0.00265 mol MnO4- x 5 = 0.01325 mol
Step 4: Convert moles into mass of iron
Mass of iron = 0.01325 mol x 55.85 gmol-1 = 0.740 g
Step 5: find the percentage of iron in the tablet
∴ % Fe in the tablet = (0.740/ 2.25) x 100 = 32.9%
Cr2O7- (aq) + 14H+ (aq) + 6e- → 2Cr3+ (aq) + 7H2O (l)
orange green


The steps in producing a balanced redox equation between iron(II) and dichromate(VI)
Always show your working in redox titration problems as marks can be awarded for the steps even if the final answer is wrong.
转载自savemyexams
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