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 SO42- would be entered as SO4^2-, with the "^" character indicating a superscript.
Leave a space between components.

I3- (aq) + 2e- → 3 I- (aq) ξo = 0.55
Tl3+ (aq) + 2e- → Tl+ (aq) ξo = 1.24

Answer 1

Answer:

3I^-/I3^-//Tl3^+//Tl^+

Explanation:

Based on the given reduction potentials, the half reaction equation having greater positive value will be the cathode while the half reaction equation having less positive reduction potential will be the anode.

This implies that this equation:

Tl3+ (aq) + 2e- → Tl+ (aq) ξo = 1.24 occurs at the cathode

While:

3 I- (aq) → I3- (aq) + 2e- ξo = -0.55 occurs at the anode

Cell notation,

Anode//cathode

Hence for this system under consideration:

3I^-/I3^-//Tl3^+//Tl^+