Given the following pair of half-reactions, fill in the line notation for a stan

Question

Given the following pair of half-reactions, fill in the line notation for a standard cell using these two half-reactions. Since it is a standard cell, all concentrations will be 1 M, so do not enter concentrations. You have solid platinum (Pt) electrodes to use, where necessary. Do not include the physical state of the components, i.e., solid platinum would just be Pt. For subscript and superscripts use the model: Ag^+ would be Ag^+ and SO_4^2- would be entered as SO4^2-, with the "^" character indicating a superscript. Leave a space between components. Mn^2+ (aq) + 2e^- rightarrow Mn (s) xi = -1.18 Fe^2+ (aq) + 2e^- rightarrow Fe (s) xi = -0.44 All reactants and products must be included. The anode, where oxidation takes place, is written on the left. solid anode | solution in anode || solution in cathode | solid cathode

Explanation / Answer

The more negative the standard reduction potential, the more difficult it is to reduce the metal ion in the solution to the consequent metal; therefore, it is easier to oxidize the metal to the corresponding metal ion. Since, Mn^2+,Mn has a more negative reduction potential, therefore, it is easier to oxidize Mn to Mn^2+. Therefore, Fe^2+ is reduced to Fe. The solid support (Pt electrode) doesn’t take part in the reaction. Therefore, the reaction taking place is

Mn (s) + Fe^2+ (aq) -------> Mn^2+ (aq) + Fe (s)

E0 = E0red + E0ox = (-0.44 V) + (1.18 V) (the oxidation potential is the reverse of the reduction potential)

===> E0 = 0.74 V

The cell is represented as

Pt MnMn^2+Fe^2+FePt

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