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Question
Consider the following standard electrode potential values:
\[\ce{Fe^{3+}_{ (aq)} + e^- -> Fe^{2+}_{ (aq)}}\], E0 = +0.77 V
\[\ce{MnO^{-4}_{ (aq)} + 8H^+ + 5e^- -> Mn^{2+}_{ (aq)} + 4H2O_{(l)}}\], E0 = +1.51 V
What is the cell potential for the redox reaction?
Options
−2.28 V
−0.74 V
+0.74 V
+2.28 V
Solution
+0.74 V
Explanation:
Cell potential (E) = \[\ce{E_{cathode} − E_{anode}}\]
= 1.51 − 0.77
= 0.74 V
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