By carrying out several titrations, the average value of the volume of EDTA used for reaching the end-point can be calculated, together with the known molarity of EDTA and nickel (II) ions (both 0.1M), the no. of moles of EDTA used and that of nickel (II) ions can be found, by considering the mole ratio between them, the formula of Nickel-EDTA complex can be obtained.
- Results:
Average volume of 0.1M EDTA used = (25.3+24.6+24.8) / 3 = 24.9 dm3
No. of moles of EDTA used = (24.9/1000) x 0.1 = 2.49x10-3 mol
No. of moles of Ni2+ used = (25.0/1000) x 0.1 = 2.59x10-3 mol
Therefore, the mole ration of EDTA to Ni2+ ~ 1:1
The formula of the Nickel-EDTA complex is then [Ni(EDTA)]
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Discussion:
Questions
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Zn reduced part of VO2+ (yellow) into VO2+ (blue), the mixing of the two species give a resulting green solution.
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Yellow→ yellowish green → bluish green → blue → dark blue → bluish green → dark green → grey → violet. It is because Zn reduce VO2+ gradually to VO2+, and then V3+ and finally V2+
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The pale yellow colour of the solution gradually turns orange → brown, after the addition of thiosulphate, it turns yellowish green, bluish green and finally pale blue. It is because before the addition of thiosulphate, VO2+ was reduce to VO2+ while at the same time the iodide ions added was oxidized to iodine solid, thus showing a brown colour. After thiosulphate solution was added, it reduced the iodine back into iodide and hence the real colour of VO2+ (blue) appears again.
- It removes the iodine solid in the mixture by reducing it back to iodide ions with the following equation:
I2(s) + 2S2O32-(aq) S4O62-(aq)+ 2I-(aq)
As a result, the interference of colour due to the black iodine solid can be removed, and hence the real colour of VO2+ (blue) appears again
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It is because zinc is a stronger reducing agent (with a reduction potential of +0.76V) than iodide (-0.54V), therefore Zn can reduce vanadium in VO2+ to the lowest oxidation state (+2) while iodide can only reduce it to +3
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The solution turns blue and gas of characteristic pungent choking smell evolved. It does not completely correspond to the result obtained by predicting using the E0 values.
Considering the following reactions:
VO2+(aq) + 2H+(aq) + e- VO2+(aq) + H2O(l) E0 = +1.00 V
SO42-(aq) + 4H(aq)++ 2e- H2SO3(aq) + H2O(l) E0 = +0.17 V
The overall E0 = +1.00 - 0.17 = +0.83 V > 0
Therefore, sulphite ion is able to reduce VO2+ to VO2+
Considering the following reactions:
VO2+(aq) + 2H+(aq) + e- V3+(aq) + H2O(l) E0 = +0.34 V
SO42-(aq)+ 4H+(aq)+ 2e- H2SO3(aq) + H2O(l) E0 = +0.17 V
The overall E0 = +0.34 - 0.17 = +0.17 V > 0
Therefore, theoretically sulphite ion is able to reduce VO2+ to V3+
However, this does not match the experiment result that the solution remains blue even further addition of sulphite. This may be due to the too low magnitude of the overall E0 (+0.17 V) which is in fact not very favourable for this further reduction.
- Considering the following reactions:
V2+(aq) V3+(aq) + e- E0 = +0.26 V
SO42-(aq) + 4H+(aq)+ 2e- H2SO3(aq) + H2O(l) E0 = +0.17 V
The overall E0 = +0.26 + 0.17 = +0.43 V > 0
Therefore, sulphuric acid is able to reduce V2+ to V3+
BUT Considering the following reactions:
VO2+(aq) + 2H+(aq) + e- V3+(aq) + H2O(l) E0 = +0.34 V
SO42-(aq)+ 4H+(aq)+ 2e- H2SO3(aq) + H2O(l) E0 = +0.17 V
The overall E0 = + 0.26 - 0.34 = -0.08 V
Therefore, sulphuric acid is unable to further reduce V3+ to VO2+
Therefore, the suitable oxidizing agent is concentrated sulphuric acid.
Sources of Error
The zinc powder added in the third step may not be of sufficient amount to reduce VO2+ to the final V2+
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Conclusion:
There are 4 oxidation states of vanadium, which are VO2+(+5), VO2+(+4),
V3+(+3), V2+(+2) respectively.
Zinc is able to reduce VO2+ into V2+,
Permanganate is able to oxidize V2+ into VO2+.
Sulphite is able to reduce VO2+ to VO2+.
VO2+ is able to oxidize V2+ to V3+.
Iodide is able to reduce VO2+ to V3+.
Finally concentrated sulphuric acid is able to oxidize V2+ to V3+
