JEE MAIN - Chemistry (2014 (Offline) - No. 7)
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 :
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 :
– 0.33 V; the reaction will not occur
– 0.33 V; the reaction will occur
– 2.69 V; the reaction will not occur
– 2.69 V; the reaction will occur
Explanation
(a)$$\,\,\,\,\,$$ $$M{n^{2 + }} + 2{e^ - } \to Mn;\,\,{E^ \circ } = - 1.18V;\,...\left( i \right)$$
(b) $$\,\,\,\,\,$$ $$M{n^{3 + }} + e \to M{n^{2 + }};\,\,{E^ \circ } = - 1.51V;\,...\left( {ii} \right)$$
Now multiplying equation $$(ii)$$ by two and subtracting from equation $$(i)$$
$$3M{n^{2 + }} \to M{n^ + } + 2M{n^{3 + }};$$
$${E^ \circ } = {E_{{O_X}}} + {E_{{\mathop{\rm Re}\nolimits} d.}} = - 1.18 + \left( { - 1.51} \right) = - 2.69V$$
[ $$-ve$$ value of $$EMF$$ (i.e., $$\Delta G = + ve$$) shows that the reaction is non-spontaneous ]
(b) $$\,\,\,\,\,$$ $$M{n^{3 + }} + e \to M{n^{2 + }};\,\,{E^ \circ } = - 1.51V;\,...\left( {ii} \right)$$
Now multiplying equation $$(ii)$$ by two and subtracting from equation $$(i)$$
$$3M{n^{2 + }} \to M{n^ + } + 2M{n^{3 + }};$$
$${E^ \circ } = {E_{{O_X}}} + {E_{{\mathop{\rm Re}\nolimits} d.}} = - 1.18 + \left( { - 1.51} \right) = - 2.69V$$
[ $$-ve$$ value of $$EMF$$ (i.e., $$\Delta G = + ve$$) shows that the reaction is non-spontaneous ]
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