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.
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