Verified Solution • 2025–26 Session moderate

QJEEM14Ch73e5: JEE Main 2014 Chemistry - Class 12 JEEM14Ch73e5 Solution: Given below are the half-cell reactions: Mn 2+ + 2e - Mn; E o = …

Given below are the half-cell reactions:

Mn2+ + 2e- →\to Mn; Eo = -1.18 V

2(Mn3+ + e- →\to Mn2+); Eo = +1.51 V

The Eo for 3Mn2+ →\to Mn + 2Mn3+ will be :

Multiple Choice Options

  • A – 0.33 V; the reaction will not occur
  • B – 0.33 V; the reaction will occur
  • C – 2.69 V; the reaction will not occur
  • D – 2.69 V; the reaction will occur

Step-by-Step Verified Solution

(a)      \,\,\,\,\, Mn2++2e−→Mn;  E∘=−1.18V; ...(i)M{n^{2 + }} + 2{e^ - } \to Mn;\,\,{E^ \circ } = - 1.18V;\,...\left( i \right)

(b)      \,\,\,\,\, Mn3++e→Mn2+;  E∘=−1.51V; ...(ii)M{n^{3 + }} + e \to M{n^{2 + }};\,\,{E^ \circ } = - 1.51V;\,...\left( {ii} \right)

Now multiplying equation (ii)(ii) by two and subtracting from equation (i)(i)

3Mn2+→Mn++2Mn3+;3M{n^{2 + }} \to M{n^ + } + 2M{n^{3 + }};

E∘=EOX+ERed.=−1.18+(−1.51)=−2.69V{E^ \circ } = {E_{{O_X}}} + {E_{{\mathop{\rm Re}\nolimits} d.}} = - 1.18 + \left( { - 1.51} \right) = - 2.69V

[ −ve-ve value of EMFEMF (i.e., ΔG=+ve\Delta G = + ve ) shows that the reaction is non-spontaneous ]

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