3.2 Redox - WJEC
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3.2 Redox Ion-electron half equations The following half equations need to be recalled. Acidified manganate(VII) ions MnO 4 – (aq) + 8H + (aq) + 5e – ⇌ Mn 2+ (aq) + 4H 2 O(l) colour change - purple to colourless Acidified dichromate ions Cr 2 O 7 2– (aq) + 14H + (aq) + 6e – ⇌ 2Cr 3+ (aq) + 7H 2 O(l) colour change - orange to green Iodine to iodide ions I 2 (aq) + 2e – ⇌ 2I – (aq) colour change - brown to colourless Thiosulfate ions 2S 2 O 3 2– (aq) ⇌ S 4 O 6 2– (aq) + 2e – Aqueous dichromate ions and iron(II) ions Aqueous Cr 2 O 7 2– is placed in the burette and added to a conical flask where it oxidises aqueous Fe 2+ . An indicator is required. Cr 2 O 7 2– (aq) + 14H + (aq) + 6Fe 2+ (aq) ⇌ 2Cr 3+ (aq) + 7H 2 O(l) + 6Fe 3+ (aq) The interconversion of dichromate ions, Cr 2 O 7 2– and chromate(VI) ions, CrO 4 2– is NOT redox as the chromium does not change its oxidation number. The colour change is orange to yellow and vice versa. Cr 2 O 7 2– (aq) + OH – (aq) ⇌ 2CrO 4 2– (aq) + H + (aq) Aqueous thiosulfate ions and aqueous iodine Aqueous S 2 O 3 2– ions are oxidised by aqueous I 2 . The S 2 O 3 2– solution is placed in the burette; the I 2 is pipetted into the conical flask. The S 2 O 3 2– solution is run into the flask until the colour, due to I 2 , fades to a pale-yellow colour. At this point starch solution is added as an indicator, turning the mixture dark blue. The end point is when the blue colour is decolourised. I 2 (aq) + 2S 2 O 3 2– (aq) → S 4 O 6 2– (aq) + 2I – (aq) Aqueous iodide ions and aqueous copper(II) ions 2Cu 2+ (aq) + 4I – (aq) → 2CuI(s) + I 2 (aq) The blue solution loses its colour. A white precipitate of CuI forms along with a brown solution of I 2 . The I 2 can then be titrated with S 2 O 3 2– . Aqueous manganate(VII) ions and iron(II) ions The potassium manganate(VII) solution must always be standardised by titration with a standard solution of a reducing agent such as iron(II) ammonium sulfate. The aqueous MnO 4 – is placed in the burette and the Fe 2+ solution pipetted into the conical flask with an excess of aqueous sulfuric acid. These titrations require no indicator. As soon as all the Fe 2+ has been oxidised, the next drop of MnO 4 – gives a pink colour to the contents of the flask. Half equations are combined to give an overall ionic equation: Fe 2+ (aq) ⇌ Fe 3+ (aq) + e – MnO 4 – (aq) + 8H + (aq) + 5e – ⇌ Mn 2+ (aq) + 4H 2 O(l) MnO 4 – (aq) + 8H + (aq) + 5Fe 2+ (aq) ⇌ Mn 2+ (aq) + 4H 2 O(l) + 5Fe 2+ (aq) 25.00 g of FeSO 4 .7H 2 O was dissolved in water and made up to 1.00 dm 3 using excess dilute sulfuric acid. 25.0 cm 3 of this solution was titrated with aqueous KMnO 4 . 25.55 cm 3 of the KMnO 4 solution was required for complete oxidation. Calculate the concentration of the KMnO 4 solution in mol dm –3 . M r FeSO 4 .7H 2 O = 278.0 moles FeSO 4 .7H 2 O = 25.00/278.0 = 0.08933 moles Fe 2+ = 0.08933 [Fe 2+ ] = 0.08993 mol dm –3 MnO 4 – (aq) + 8H + (aq) + 5Fe 2+ (aq) ⇌ Mn 2+ (aq) + 4H 2 O(l) + 5Fe 2+ (aq) moles Fe 2+ in 25.0 cm 3 = 0.08993 × (25.0 / 1000) = 2.248 × 10 –3 moles MnO 4 – in 25.55 cm 3 = 2.248 × 10 –3 / 5 = 4.496 × 10 –4 [KMnO 4 ] = 4.496 × 10 –4 / (25.55 / 1000) = 0.0176 mol dm –3
Transcript of 3.2 Redox - WJEC
3.2 Redox
Ion-electron half equationsThe following half equations need to be recalled.
Acidif ied manganate(VII) ions
MnO4–(aq) + 8H+(aq) + 5e– ⇌ Mn2+(aq) + 4H2O(l)
colour change - purple to colourless
Acidif ied dichromate ions
Cr2O72–(aq) + 14H+(aq) + 6e– ⇌ 2Cr3+(aq) + 7H2O(l)
colour change - orange to green
Iodine to iodide ions
I2(aq) + 2e– ⇌ 2I–(aq)
colour change - brown to colourless
Thiosulfate ions
2S2O32–(aq) ⇌ S4O6
2–(aq) + 2e–
Aqueous dichromate ions and iron(II) ionsAqueous Cr2O7
2– is placed in the burette and added to a conical flask where it oxidises aqueous Fe2+.
An indicator is required.
Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) ⇌ 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
The interconversion of dichromate ions, Cr2O72– and chromate(VI) ions, CrO4
2– is NOT redox as the chromium does not change its oxidation number. The colour change is orange to yellow and vice versa.
Cr2O72–(aq) + OH–(aq) ⇌ 2CrO4
2– (aq) + H+(aq)
Aqueous thiosulfate ions and aqueous iodineAqueous S2O3
2– ions are oxidised by aqueous I2. The S2O3
2– solution is placed in the burette; the I2 is pipetted into the conical flask. The S2O3
2– solution is run into the flask until the colour, due to I2, fades to a pale-yellow colour. At this point starch solution is added as an indicator, turning the mixture dark blue. The end point is when the blue colour is decolourised.
I2(aq) + 2S2O32–(aq) → S4O6
2–(aq) + 2I–(aq)
Aqueous iodide ions and aqueous copper(II) ions
2Cu2+(aq) + 4I–(aq) → 2CuI(s) + I2(aq)
The blue solution loses its colour. A white precipitate of CuI forms along with a brown solution of I2.
The I2 can then be titrated with S2O32–.
Aqueous manganate(VII) ions and iron(II) ionsThe potassium manganate(VII) solution must always be standardised by titration with a standard solution of a reducing agent such as iron(II) ammonium sulfate. The aqueous MnO4
– is placed in the burette and the Fe2+ solution pipetted into the conical flask with an excess of aqueous sulfuric acid. These titrations require no indicator. As soon as all the Fe2+ has been oxidised, the next drop of MnO4
– gives a pink colour to the contents of the flask.
Half equations are combined to give an overall ionic equation:
Fe2+(aq) ⇌ Fe3+(aq) + e–
MnO4–(aq) + 8H+(aq) + 5e– ⇌ Mn2+(aq) + 4H2O(l)
MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) ⇌ Mn2+(aq) + 4H2O(l) + 5Fe2+(aq)
25.00 g of FeSO4.7H2O was dissolved in water and made up to 1.00 dm3 using excess dilute sulfuric acid. 25.0 cm3 of this solution was titrated with aqueous KMnO4. 25.55 cm3 of the KMnO4 solution was required for complete oxidation.
Calculate the concentration of the KMnO4 solution in mol dm–3.
Mr FeSO4.7H2O = 278.0
moles FeSO4.7H2O = 25.00/278.0 = 0.08933
moles Fe2+ = 0.08933
[Fe2+] = 0.08993 mol dm–3
MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) ⇌ Mn2+(aq) + 4H2O(l) + 5Fe2+(aq)
moles Fe2+ in 25.0 cm3 = 0.08993 × (25.0 / 1000) = 2.248 × 10–3
moles MnO4– in 25.55 cm3 = 2.248 × 10–3 / 5 = 4.496 × 10–4
[KMnO4] = 4.496 × 10–4 / (25.55 / 1000) = 0.0176 mol dm–3
