What two factors can influence the redox potential of transition metal ions compared to standard conditions?
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Why can transition metals exist in various positive oxidation states?
Transition metals have outer d orbital electrons with similar energies, allowing multiple oxidation states to be stable in compounds and complexes.
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What is the overall equation, including state symbols, for the redox reaction between aqueous C2O42- ions and aqueous MnO4- ions in acidic conditions?
5C2O42-(aq) + 2MnO4-(aq) + 16H+(aq) ➔ 10CO2(g) + 2Mn2+(aq) + 8H2O(l)
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How can standard electrode potentials be used to predict the direction of a redox reaction?
The E⦵ values of the half-reactions involved are compared. The reaction is likely to proceed in the direction of the more positive (or less negative) E⦵ value, indicating the species that gains electrons and undergoes reduction.
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What is the change in vanadium's oxidation state in the reaction between VO2+ ions and Zn?
Vanadium is reduced from +5 (in VO2+) to +2 (in V2+).
Zinc is oxidised from 0 to +2.
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Fe3+(aq) + e- ⇌ Fe2+(aq) E⦵ = +0.77 V
MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l) E⦵ = +1.33 V
Explain which species is reduced when the two half-equations above are combined.
MnO4- is reduced to Mn2+.
This is because the E⦵ value for MnO4-/Mn2+ is more positive than the E⦵ value for Fe3+/Fe2+.
This means that Fe2+ is oxidised to Fe3+.
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What is the half-equation for the reduction of MnO4-(aq) ions to Mn2+(aq) ions in acidic conditions?
MnO4- + 8H+ + 5e- ➔ Mn2+ + 4H2O
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