Given the reaction that occurs in an electrochemical cell: zn(s) + cuso4(aq) znso4(aq) + cu(s) during this reaction, the oxidation number of zn changes from

Answer : The correct answer is oxidation number of Zn changes from 0 to +2.

Oxidation number :

It is number given to an element when it undergoes oxidation by loosing electrons . Oxidation number is equal to number of electrons donated . It is generally represented as charge .

Oxidation number of any element in a compound can be calculated as :

(Total number of positive charge) + (total number of negative charge) = total charge on compound

Where : Total positive charge = No. of atom or polyatomic molecule * positive charge on it

Total negative charge = No. of atom or polyatomic molecule * negative charge on it

Total charge on compound = Charge lying on top of compound .

Following steps can be used to find change in oxidation number of Zn :

Step 1 ) : To find oxidation number of Zn ( reactant side ) :

On reactant side Zn is present in solid state or its elemental state . It is not attached to any other atom , hence it is neutral and isolated atom. So oxidation number on Zn(s) is 0 .

Step 2: To find oxidation number of Zn in ZnSO₄ (aq)

ZnSO₄(aq) is ionic compound compound . It has two ions in it, Zn and SO₄ . Also it is a neutral compound means it has zero charge on it .

SO₄ is sulfate which has -2 charge or oxidation number on it (image attached )and there is one molecule of SO₄ ion . There is one atom of Zn which has unknown charge or oxidation number (x) .

To calculate oxidation number on Zn ion , plug values in above formula :

(No of Zn atom * charge on it ) + (No of SO₄ molecule * charge on it ) = Total charge on ZnSO₄

( 1* charge on Zn ) + ( 1* -2 ) = 0

Charge on Zn + (-2) = 0

Charge on Zn = +2

On reactant side Zn(s) has Zero charge and on product side Zn in ZnSO₄(aq) has +2 charge .

Hence the charge in oxidation number of Zn is from 0 to +2 .