Eº for Mn3+ / Mn couple is much more +ve
for Fe 3+ / Fe 2+. Why?
Answers
Answer:
(i) Mn2+ is more stable than Mn3+ because Mn+2 has exactly half-filled orbitals, while Fe+2 after losing one e– half-filled orbitals. (ii) Ce+4 achieve inert gas structure of Xenon, to require extra stability.
Why is the standard electrode potential value for Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+?
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Since we know that [del] G for a reaction is equal to -nFE…..
where
n = no. of electrons involved
F = Faraday’s constant
E = electrode potential
for a reaction to be feasible thermodynamically [del] G < 0, Hence E will be greater than zero or rather more positive…
the more thermodynamically feasible the reaction is the greater is the electrode potential…
for Mn 3+−−−−>Mn2+
Configuration of Mn3+=3d44s0
Configuration of Mn2+=3d54s0 (here stable half filled configuration is being formed)
So, this reaction is more favourable than…
Cr 3+−−−−>Cr2+
Configuration of Cr3+=3d34s0 (here the t^2g orbitals are fully filled and hence this is more stable)
Configuration of Cr2+=3d44s0
Since reactant is more stable the reaction is unfavourable or E<0
Fe 3+−−−−>Fe2+
Configuration of Fe3+=3d54s0 (here stable half filled configuration is present)
Configuration of Fe2+=3d64s0
Since reactant is more stable the reaction is unfavourable or E<0
Hence, the E value of first reaction will be greater than the other two..