For an oxidation M(II)→M(III), standard reduction potentials E°[M(III)/M(II)] are +1.30 V in ligand set A and −0.20 V in ligand set B. The oxidant O has E°[Oₒₓ/ O re d ] = +0.80 V. All potentials re

Question

For an oxidation M(II)→M(III), standard reduction potentials E°[M(III)/M(II)] are +1.30 V in ligand set A and −0.20 V in ligand set B. The oxidant O has E°[Oₒₓ/ O re d ​ ] = +0.80 V. All potentials refer to the same medium and temperature. Which statements are correct?

TestHub · Accepted Answer

Oxidising M(II) gives E° ce ll = E°(oxidant)−E°(metal reduction). Thus A: 0.80−1.30 = −0.50 V; B: 0.80−(−0.20) = +1.00 V. The lower metal reduction potential means M(III) is harder to reduce and relatively stabilised in B. No inference about magnetic moments is needed.

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