Manganese ions (Mn^{2+}) can be oxidised by p | vedclass

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(A) To balance the redox reaction,the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reducti

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$2.5$

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(A) To balance the redox reaction,the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction.Reduction half-reaction: $S_2O_8^{2-} + 2e^- \longrightarrow 2SO_4^{2-}$ (Gain of $2e^-$)Oxidation half-reaction: $Mn^{2+} + 4H_2O \longrightarrow MnO_4^- + 8H^+ + 5e^-$ (Loss of $5e^-$)To equate the electrons,multiply the reduction half-reaction by $5$ and the oxidation half-reaction by $2$:$5S_2O_8^{2-} + 10e^- \longrightarrow 10SO_4^{2-}$$2Mn^{2+} + 8H_2O \longrightarrow 2MnO_4^- + 16H^+ + 10e^-$From the balanced equation,$2 \ moles$ of $Mn^{2+}$ require $5 \ moles$ of $S_2O_8^{2-}$.Therefore,$1 \ mole$ of $Mn^{2+}$ requires $5/2 = 2.5 \ moles$ of $S_2O_8^{2-}$.

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$2 MnO_4^{-} + bI^{-} + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^{-}$If the above equation is balanced with integer coefficients,the value of $z$ is . . . . . .

For the redox reaction $MnO_4^- + C_2O_4^{2-} + H^{+} \to Mn^{2+} + CO_2 + H_2O$,the correct coefficients of the reactants for the balanced reaction are:$MnO_4^-$ : $C_2O_4^{2-}$ : $H^{+}$