Question

Course Contents > Electrochem IV-Graded HomERO 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.2 would be entered as s042, with the "A" character indicating a superscript. Leave a space between components. All reactants and products must be included The anode, where oxidation takes place, is written on the left. solid anodelsolution in anodel Isolution in cathodelsolid cathode Submit Answer Tries 7/45 Previous Tries My general preferences on what Mark NEW p Export NEW Matthew James Larson Reply (Mon Mar 11 12:59:01 pm 2019 (CDT)) How do we do this problem? Even if we use the platinum, that's only one solid, and within the scope of this problem we have no way to f solid form of either compound, but the hint text says we should have a solid form of the anode and cathode. NEW Re: Deiah C Kazes Reply (Mon Mar 11 06:24:47 pm 2019 (CDT)) When I did this problem, I disregarded the "solid platinum" information and worked with the two half reactions given. Both states are each reaction (solid and aqueous), so our job is to assign the correct formula to the correct cell without including their physical states instance, placing "Mn" on the correct side of the two bars in the middle, and on the correct side of the single bar.) NEW Raafac Zaki Reply (Tue Mar 12 04:54:08 pm 2019 (CDT)) how do you do this with Cobalt? I have an aqueous cobalt solutions as product and reactant Tne here to search

Answer 1

The half equations are

I₃^- (aq) + 2e^- -------->3 I^- (aq)       ξ⁰ = 0.55 V

Tl³⁺ (aq) + 2e^- --------> Tl⁺ (aq)    ξ⁰ = 1.24 V

The more positive is the standard reduction potential of a system, the greater is the tendency of the species to be reduced. Since Tl³⁺/Tl⁺ has a higher reduction potential than I₃^-/I^-, hence, Tl³⁺ will be preferentially reduced and I^- will be oxidized to I₃^-.

The reduction takes place at the cathode and oxidation takes place at the anode. Pt provides the solid support.

While writing the cell notation, the oxidized species is written first followed by the reduced species and the two are separated by a comma. Therefore, the cell notation is

Pt│I^3-, I^-││Tl^3+, Tl^+│Pt