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Electrochemistry question

2022 · Q111
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Electrochemistry question

2022 · Q111

NEETChemistryElectrochemistryMCQ+4 / −1

Given below are half cell reactions:

MnO4−+8H++5e−→Mn2++4H2OMnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}OMnO4−​+8H++5e−→Mn2++4H2​O,

EMn2+/MnO4−o=−1.510 VE_{M{n^{2 + }}/MnO_4^ - }^o = - 1.510\,VEMn2+/MnO4−​o​=−1.510V

12O2+2H++2e−→H2O{1 \over 2}{O_2} + 2{H^ + } + 2{e^ - } \to {H_2}O21​O2​+2H++2e−→H2​O

EO2/H2Oo=+1.223 VE_{{O_2}/{H_2}O}^o = + 1.223\,VEO2​/H2​Oo​=+1.223V

Will the permanganate ion, MnO4−MnO_4^ -MnO4−​ liberate O2 from water in the presence of an acid?

  1. A
    Yes, because EcelloE_{cell}^oEcello​ = + 0.287 V
  2. B
    No, because EcelloE_{cell}^oEcello​ = −-−0.287 V
  3. C
    Yes, because EcelloE_{cell}^oEcello​ = + 2.733 V
  4. D
    No, because EcelloE_{cell}^oEcello​ = −-− 2.733 V
View written solutionFree

Correct answer: A

$\bullet$ $$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$$ ..... (i)

$$E_{MnO_4^ - /M{n^{2 + }}}^o = - E_{M{n^{2 + }}/MnO_4^ - }^o = 1.51\,V$$

$\bullet$ $${H_2}O \to {1 \over 2}{O_2} + 2{H^ + } + 2{e^ - }$$ ..... (ii)

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

Using 2 $\times$ (i) + 5 $\times$ (ii), net cell reactions is

$$2MnO_4^ - + 6{H^ + } \to 2M{n^{2 + }} + {5 \over 2}{O_2} + 3{H_2}O$$

$$E_{cell}^o = E_C^o - E_A^o = E_{MnO_4^ - /M{n^{2 + }}}^o - E_{{O_2}/{H_2}O}^o = 1.51 - 1.223 = 0.287\,V$$

Since $E_{cell}^o > 0$, therefore net cell reaction is spontaneous and so $MnO_4^ - $ liberate O2 from H2O in presence of an acid.

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