What is the value for the e.m.f. of the cell represented by Pt(s)|OH-(aq)|O2(g)||Fe2+(aq)|Fe(s)?

To determine the value of the electromotive force (e.m.f.) of the electrochemical cell represented by the notation Pt(s)|OH–(aq)|O2(g)||Fe2+(aq)|Fe(s), it’s necessary to analyze the half-reactions occurring in the cell and their standard electrode potentials.

In this cell, we have two half-reactions occurring at different electrodes:

1. The reduction half-reaction at the cathode involves the reduction of oxygen gas (O2) in the presence of hydroxide ions (OH–). The standard reduction potential for this half-reaction can be referenced from standard tables, which typically lists the standard reduction potential for O2 + 4e– + 2H2O → 4OH– is +0.40 V.

2. The oxidation half-reaction at the anode is the oxidation of ferrous ions (Fe2+) to solid iron (Fe). The standard oxidation potential for the ferrous ion is the negative of the standard reduction potential for Fe2+ + 2e– → Fe, which is commonly found to be about -0.44 V.

Next, we calculate the standard cell potential (e.m.f.) using the standard reduction potentials of the two