Question

1) Write the reaction on anode?
2) Write the reaction on cathode?
3) What is the formula of rust?
4) What is the formula of water?
5) Which element take part in the oxidation process​

Answer 1

The half-reaction on the anode, where oxidation occurs, is Zn(s) = Zn2+ (aq) + (2e-). The zinc loses two electrons to form Zn2+. The half-reaction on the cathode where reduction occurs is Cu2+ (aq) + 2e- = Cu(s). Here, the copper ions gain electrons and become solid

Rust is apparently a hydrated form of iron(III)oxide. The formula is approximately Fe2O3•32H2O, although the exact amount of water is variable. (Note that this is about halfway between iron(III) hydroxide, Fe(OH)3 or ½{Fe2O3•3H2O], and anhydrous Fe2O3).

formula of water => H2O

The oxidation state of a pure element is always zero. The oxidation state for a pure ion is equivalent to its ionic charge. In general, hydrogen has an oxidation state of +1, while oxygen has an oxidation state of -2.

Answer 2

Explanation:

The half-reaction on the anode, where oxidation occurs, is Zn(s) = Zn2+ (aq) + (2e-). The zinc loses two electrons to form Zn2+. The half-reaction on the cathode where reduction occurs is Cu2+ (aq) + 2e- = Cu(s). Here, the copper ions gain electrons and become solid copper.

this is the answer of 1

The half-reaction on the cathode where reduction occurs is Cu2+ (aq) + 2e- = Cu(s). Here, the copper ions gain electrons and become solid copper. The entire reaction can be written by combining both half-reactions: Zn(s) + Cu2+ (aq) = Zn2+ (aq) + Cu(s).

this is the answer of 2

Rust is apparently a hydrated form of iron(III)oxide. The formula is approximately Fe2O3•32H2O, although the exact amount of water is variable. (Note that this is about halfway between iron(III) hydroxide, Fe(OH)3 or ½{Fe2O3•3H2O], and anhydrous Fe2O3)

this is the answer of 3

H20

this is the answer of 4

oxygen atom

this is the answer of 5