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Electrochemistry
Electrochemical Cells and Feasibility of Reactions
1 question from 1 paper, each with its answer.
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Concepts tested
- cell potential from half cell potentials
All Questions
2022 1 question
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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?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.