Combine the two half-reactions that give the spontaneous cell reaction with the smallest e∘. fe2+(aq)+2e−→fe(s) e∘=−0.45v i2(s)+2e−→2i−(aq) e∘=0.54v cu2+(aq)+2e−→cu(s) e∘=0.34v

The spontaneous cell reaction having smallest is

Explanation:

We are given:

The substance having highest positive potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

Oxidation half reaction:    

Reduction half reaction:    

Thus, this cell will not give the spontaneous cell reaction with smallest

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

Oxidation half reaction:    

Reduction half reaction:  

Thus, this cell will give the spontaneous cell reaction with smallest

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

Oxidation half reaction:    

Reduction half reaction:    

Thus, this cell will not give the spontaneous cell reaction with smallest

Hence, the spontaneous cell reaction having smallest is

The correct answer would be the last option. in a neutral solution, the concentration of hydrogen ions is equal to the concentration of hydroxide ions. the ph of this solution would be 7. it is neither acidic nor basic. the hydroxide and hydronium ions are equal in amount and are in equilibrium.