Respuesta :
Answer: By looking at the standard reduction potentials, we can say that bromine will always get reduced in the following reaction.
Explanation:
For the given chemical reaction:
[tex]I_2(s)+2Br^-(aq.)\rightarrow 2I^-(aq.)+Br_2(l);\Delta G^o=1.1\times 10^5J[/tex]
Here, iodine is getting reduced because it is gaining electrons and bromine is getting oxidized because it is loosing electrons.
We know that:
[tex]E^o_{(I_2/I^-)}=0.53V\\E^o_{(Br_2/Br^-)}=1.07V[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
[tex]E^o_{cell}=0.53-1.07=-0.54V[/tex]
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.
By looking at the standard electrode potential of the cells, the substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.
Therefore, for the given reaction, bromine must undergo reduction reaction and act as cathode. And, iodine must undergo oxidation reaction and act as anode.
What is spontaneous reaction?
A reaction is said to be spontaneous when it is able to occur on its own with minimal or no energy input externally. If we look at this reaction, it is not sponteanous because the change in free energy for the reaction is positive (ΔG∘=1.1×105J).
This could be understood from the fact that bromine is above iodine in the electrochemical series and can not esily be displaced by iodine except by an aditional input of energy to drive the reaction.
Learn more about spontaneous reaction: https://brainly.com/question/13232104?
