A Latimer diagram for a chemical element is a summary of the standard electrode potential data for that element. In a Latimer diagram, the form of the element with the highest oxidation state is on the left, with successively lower oxidation states to the right. A Latimer diagram for manganese at $$\(\mathrm{pH}=0\)$$ is shown. The diagram shows that the standard electrode potential for the reduction of $$\(\mathrm{MnO}_{4}^{-}\)$$to $$\(\mathrm{MnO}_{2}\)$$, in acidic conditions, is +1.69 V . $$\[ \mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightleftharpoons \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \quad E^{\ominus}=+1.69 \mathrm{~V} \]$$ Justify the assignment of the oxidation state of +5 to manganese in $$\(\mathrm{H}_{3} \mathrm{MnO}_{4}\)$$ using oxidation numbers. (1) ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ (ii) Write an equation for the reaction of $$\(\mathrm{H}_{3} \mathrm{MnO}_{4}\)$$ in acidic solution to give ions containing manganese(VI) and manganese(IV). Use the Latimer diagram to obtain the formulae of the ions produced. State symbols are not required. (2) (iii) Deduce whether or not this disproportionation reaction is thermodynamically feasible by calculating $$\(E^{\ominus}\)$$ for the reaction. (2)

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