August 3rd, 2007<\/p>\n
Manganese (VI) is usually in the form of the MnO4<\/sub>2-<\/sup> ion. This ion disproportionates in acid solution but not in base. What is going on here?<\/p>\n The Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation – it is a reaction):<\/p>\n 3MnO4<\/sub>(2-) + 4H+ \u2013> MnO2<\/sub> + 2MnO4<\/sub>(-) + 2H2<\/sub>O<\/p>\n The Mn(VI) state cannot do this under alkaline conditions, and the alkaline disproportionation reaction (theoretical) would be:<\/p>\n 3MnO4<\/sub>(2-) + 2H2<\/sub>O \u2013> MnO2<\/sub> + 2MnO4<\/sub>(-) + 4OH-<\/p>\n This is now in direct competition with the reverse reaction (as all reactions are) which we know DOES occur.<\/p>\n However, what we are really comparing is the Gibbs free energy of the two disproportionation equations. In the case of acid conditions:<\/p>\n 3MnO4<\/sub>(2-) + 4H+ \u2013> MnO2<\/sub> + 2MnO4<\/sub>(-) + 2H2<\/sub>O<\/p>\n The Gibbs Free energy makes the forward reaction feasible. In the case of alkaline conditions however, Gibbs Free energy makes the backward reaction equally feasible.<\/p>\n 3MnO4<\/sub>(2-) + 2H2<\/sub>O <\u2013 MnO2<\/sub> + 2MnO4<\/sub>(-) + 4OH-<\/p>\n This can be predicted\u00a0by reference to the\u00a0electrode potential of the half equations involved. If you break down each disporportionation into two half-equations and compare the redox potentials using E = E(red) – E(ox).<\/p>\n OK, let\u2019s do it! 3MnO4<\/sub>(2-) + 4H+ \u2013> MnO2<\/sub> + 2MnO4<\/sub>(-) + 2H2<\/sub>O<\/p>\n This disproportionation can be though of as two half equations: Calculating E\u00ba = E(reduced state) – E(oxidised state) = +2.26V – +0.56V = + 1.70V 3MnO4<\/sub>(2-) + 2H2<\/sub>O \u2013> MnO2<\/sub> + 2MnO4<\/sub>(-) + 4OH-<\/p>\n This disproportionation can be though of as two half equations: Calculating E\u00ba = E(red) – E(ox) = 0.67 – 0.56 =\u00a0 +0.11V August 3rd, 2007 Manganese (VI) is usually in the form of the MnO42- ion. This ion disproportionates in acid solution but not in base. What is going on here? The Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation – it is a reaction): 3MnO4(2-) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[19,24,39,40,48,60,63,62],"class_list":["post-7","post","type-post","status-publish","format-standard","hentry","category-inorganic","tag-acid","tag-basic","tag-disproportionate","tag-disproportionation","tag-gibbs-free-energy","tag-manganese-vi","tag-mno4","tag-mno42"],"_links":{"self":[{"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/posts\/7","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/comments?post=7"}],"version-history":[{"count":0,"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/posts\/7\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/media?parent=7"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/categories?post=7"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.ibchem.com\/blog\/wp-json\/wp\/v2\/tags?post=7"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014
\nAcidic conditions:<\/p>\n
\n1. MnO4<\/sub>(2-) + 4H+ + 2e \u2013> MnO2<\/sub> + 2H2<\/sub>O \u2026\u2026\u2026.. E\u00ba = +2.26V ( the Manganate ion is reduced)
\n2. MnO4<\/sub>(2-) \u2013> MnO4<\/sub>(-) + 1e \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.E\u00ba = +0.56V ( the Manganate ion is oxidised<\/p>\n
\nThis is a large positive value (remember that any E\u00ba value greater than 0.3v means that the reaction is spontaneous as shown) so the forward reaction proceeds.
\nAlkaline conditions:<\/p>\n
\nMnO4<\/sub>(2-) + 2H2<\/sub>O + 2e \u2013> MnO2<\/sub> + 4OH- \u2026\u2026\u2026.. E\u00ba = +0.67V (reduction)
\nMnO4<\/sub>(2-) \u2013> MnO4<\/sub>(-) + 1e\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026E\u00ba = +0.56V (oxidation)<\/p>\n
\nAlthough this is a positive value, it is very small (under 0.3V) indicating that an equilibrium will be established (under standard conditions) and that the reaction will proceed in the forward direction by only an immeasurable amount, and as the equation is in equilibrium, by Le Chatelier, addition of base will drive the equilibrium in the reverse direction.<\/p>\n","protected":false},"excerpt":{"rendered":"