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

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(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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