Question

What substance is reduced in the following redox reaction?

Answer 1

Well, `"cupric ion"` is REDUCED....

Explanation:

A species undergoes reduction when its oxidation number is `"REDUCED"`, i.e. it formally accepts electrons...

and so...`Cu^(2+) + 2e^(-) rarr Cu(s)darr`

i.e. `Cu(II+) rarr Cu(0)`

The electrons are proposed to come from somewhere, i.e. from zinc metal, the which is conceived to LOSE electrons, and is `"OXIDIZED"`.

`Zn(s) rarr Zn^(2+) + 2e^(-)`

And to write the electrochemical cell, we simply add the individual redox equations in such a way that the electrons, virtual particles of convenience are eliminated...

`Cu^(2+) + Zn(s)+ cancel(2e^(-)) rarr Cu(s)darr+Zn^(2+) + cancel(2e^(-))`

...to give finally...

`Cu^(2+) + Zn(s)rarr Cu(s)darr+Zn^(2+)`

This is the basis of the original galvanic cell....