JEE Main 2014 · 6 Apr · Shift 1 · Chemistry Q48

← JEE Main 2014 · 6 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 48 of 90

Given below are the half - cell reactions :

$\ce{Mn^2+ + 2e- -> Mn}$ ; $E^\circ=-1.18$ V

$2\left(\ce{Mn^3+ + e- -> Mn^2+}\right)$ ; $E^\circ=+1.51$ V

The $E^\circ$ for $\ce{3Mn^2+ -> Mn + 2Mn^3+}$ will be :

-2.69

V ; the reaction will not occur

-2.69

V ; the reaction will occur

-0.33

V ; the reaction will not occur

-0.33

V ; the reaction will occur

Answer and worked solution

Official answer: A

$\ce{3Mn^2+ -> Mn + 2Mn^3+}$ pairs reduction of $\ce{Mn^2+}$ to Mn ( $-1.18$ V) with oxidation of $\ce{Mn^2+}$ to $\ce{Mn^3+}$ (reverse of $+1.51$ V):
$E^\circ=-1.18-1.51=-2.69\text{ V}.$
$E^\circ<0$ means $\Delta G^\circ>0$ , so the reaction will not occur.

Topics:Galvanic cells and EMF Gibbs energy and EMF

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JEE Main 2014 · 6 Apr · Shift 1, Chemistry question 48