Given below are half cell reactions: $MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$, $E_{M{n^{2 + }}/MnO_4^ - }^o = - 1.510\,V$ ${1 \over 2}{O_2} + 2{H^ + } + 2{e^ - } \to {H_2}O$ $E_{{O_2}/{H_2}O}^o = + 1.223\,V$ Will the permanganate ion, $MnO_4^ - $ liberate O2 from water in the presence of an acid?

Question

Given below are half cell reactions:

$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$,

$E_{M{n^{2 + }}/MnO_4^ - }^o = - 1.510\,V$

${1 \over 2}{O_2} + 2{H^ + } + 2{e^ - } \to {H_2}O$

$E_{{O_2}/{H_2}O}^o = + 1.223\,V$

Will the permanganate ion, $MnO_4^ - $ liberate O₂ from water in the presence of an acid?

Yes, because $E_{cell}^o$ = + 0.287 V

No, because $E_{cell}^o$ = $-$0.287 V

Yes, because $E_{cell}^o$ = + 2.733 V

No, because $E_{cell}^o$ = $-$ 2.733 V

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