Given below are half cell reactions: MnO₄⁻+8H⁺+5e⁻→ Mn²⁺+4H₂O E°_Mn²⁺/MnO₄⁻=-1.510 V 1/2O₂+2H⁺+2e⁻→ H₂O… | Electrochemical Cells and Feasibility of Reactions

Looking for classes? Ksquare Career Institute, Bengaluru →

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 $E^\circ_{cell}=+2.733\ V$

The answer is B.

Solution

Cathode (reduction): $2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$; $E^\circ_{RP} = 1.510$ V Anode (oxidation): $5H_2O \rightarrow \frac{5}{2}O_2 + 10H^+ + 10e^-$; $E^\circ_{RP} = 1.223$ V Net: $2MnO_4^- + 6H^+ \rightarrow 2Mn^{2+} + \frac{5}{2}O_2 + 3H_2O$ $E^\circ_{cell} = (SRP)_{cathode} - (SRP)_{anode} = 1.510 - 1.223 = 0.287$ V Since $E^\circ_{cell}$ is positive, yes – permanganate can liberate $O_2$ from water in acid.

Concept cell potential from half cell potentials

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?

Solution

Electrochemistry in past papers

Practice

Free resources

Platform

At the institute