Q. Given below are half cell reactions: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ $E^\circ_{Mn^{2+}/MnO_4^-} = -1.510\ V$ $\frac{1}{2}O_2 + 2H^+ + 2e^- \rightarrow H_2O$ $E^\circ_{O_2/H_2O} = +1.223\ V$ Will the permanganate ion, $MnO_4^-$ liberate $O_2$ from water in the presence of an acid?
A) No because $E^\circ_{cell} = -2.733\ V$
B) Yes, because $E^\circ_{cell} = +0.287\ V$
C) No, because $E^\circ_{cell} = -0.287\ V$
D) Yes, because $E^\circ_{cell} = +2.733\ V$
Explanation
[Solution] $E^\circ_{MnO_4^-/Mn^{2+}} = +1.510$ V. $E^\circ_{cell} = 1.510 - 1.223 = +0.287$ V, which is positive, so $MnO_4^-$ can oxidise water to $O_2$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.