For the cell reaction –

(S2O8)2- (aq) + 2I- ---> 2(SO4)2- + I2 (s)

What will be the CATHODE and ANODE reaction?

In the given cell reaction, we have the reaction of persulfate ions (S2O8)2- with iodide ions (I-) to produce sulfate ions (SO4)2- and iodine (I2). To determine the cathode and anode reactions, we first need to identify the oxidation and reduction processes occurring in the reaction.

Identifying Oxidation and Reduction

In electrochemical cells, oxidation occurs at the anode and reduction occurs at the cathode. Oxidation is the loss of electrons, while reduction is the gain of electrons. Let's analyze the species involved:

• (S2O8)2-: This species is reduced to sulfate ions (SO4)2-.
• I-: This species is oxidized to iodine (I2).

Oxidation Reaction at the Anode

At the anode, the iodide ions (I-) lose electrons to form iodine (I2). The half-reaction for this process can be written as:

2I- → I2 + 2e-

This shows that two iodide ions are oxidized, releasing two electrons in the process.

Reduction Reaction at the Cathode

At the cathode, the persulfate ions ((S2O8)2-) gain electrons to form sulfate ions (SO4)2-. The half-reaction for this reduction can be expressed as:

(S2O8)2- + 2e- → 2(SO4)2-

Here, the persulfate ion accepts two electrons, resulting in the formation of two sulfate ions.

Summary of Reactions

To summarize, the overall reactions occurring in the electrochemical cell can be outlined as follows:

• Anode Reaction: 2I- → I2 + 2e- (oxidation)
• Cathode Reaction: (S2O8)2- + 2e- → 2(SO4)2- (reduction)

In this cell, the anode is where oxidation occurs (the loss of electrons from iodide ions), and the cathode is where reduction takes place (the gain of electrons by persulfate ions). Understanding these processes is crucial for grasping how electrochemical cells function and how they can be applied in various chemical reactions and technologies.