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Explain the trends in M^(3+)//M^(2+) electrode potentials |
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Answer» Solution :The SCANDIUM has LOW value of `m^(3+)//M^(2+)` potentials because of its noble gas configuration. The highest value of `M^(3+)//M^(2+)` is for zinc because `Zn^(2+)` has `d^(10)` configuration which is extra stable. High value of `Mn^(3+)//Mn^(2+)` is due to half-filled `(d^(5))` orbitals in `Mn^(2+)` while low values of `Fe^(3+)//Fe^(2+)` SHOWS extra stability of `Fe^(3+) (d^(5))`. The vanadium has comparatively less negative values because of higher stability of `V^(2+)` that has half filled configuration `(t_(2g)^(3))`  The high value of `Co^(3+)//Co^(2+)` is because of high values of CFSE of `t_(2g)^(5) e^(2) [Co^(2+)]` system in a AQUEOUS medium The high values of `M^(3+)//M^(2+) " of " Mn^(3+) and Co^(3+)` indicates that they are good oxidizing agents while `Ti^(2+), V^(2+), Cr^(2+)` etc. are good reducing agents and thus these liberates `H_(2)` from acids Ex: `2Cr_((aq))^(2+) + 2H_((aq))^(+) rarr 2Cr_((aq))^(3+) + H_(2(aq)) uarr` In presence of oxidizing agents such as `HNO_(3)`, the chromium forms layer of oxide `Cr_(2)O_(3)` on its surface and hence it is passive towards oxidation. |
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