Respuesta :
Answer: The correct statement are following.
- [tex]I_{2}[/tex] is reduced at the cathode.
- In the external circuit, electrons flow from the [tex]Cu^{2+}|Cu[/tex] compartment to the [tex]I2|I^{-}[/tex] compartment.
Explanation:
An oxidizing agent is defined as a substance which readily accepts an electron and itself gets reduced in order to oxidize another substance in a chemical reaction.
So, as per the given data the reaction equations are as follows.
[tex]Cu^{2+} + 2e^{-} \rightarrow Cu(s)[/tex], [tex]E_{o}[/tex] = +0.337
[tex]I_{2}(s) + 2e^{-} \rightarrow 2I[/tex], [tex]E_{o}[/tex] = +0.54
Lower or less negative is the [tex]E_{o}[/tex] value, more easily the element or substance can lose electrons. This means that the element itself gets oxidized and reduces the other element or substance.
As per the given data, [tex]I^{-}[/tex] electrode acts as cathode and [tex]Cu^{-}[/tex] electrode acts as anode.
- Hence, [tex]I_{2}[/tex] is reduced at cathode. And, the anode compartment is [tex]Cu^{2+}|Cu[/tex] compartment.
- Also, in the external circuit electrons flow from [tex]Cu^{2+}|Cu[/tex] to the [tex]I_{2}|I^{-}[/tex] compartment.