Question

I neeed help nowww HELPPPP NOOWWWW

What will happen when a piece of magnesium metal is dropped into a beaker
containing a 1 M solution of copper(II) chloride?
Mg2+ + 2e + Mg(s) eº = -2.37 V
Cu2+ + 2e → Cu(s) E° = + 0.34 V

Answer 1

Answer: The correct option is B.

Step-by-step explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction.

The chemical species will undergo a reduction reaction if the value of standard reduction potential is more positive or less negative.

For the given half-reactions:

`Mg^(2+)+2e^-\rightarrow Mg(s);E^o_(Mg^(2+)/Mg)=-2.37V`

`Cu^(2+)(aq)+2e^-\rightarrow Cu(s);E^o_(Cu^(2+)/Cu)=+0.34V`

As the value of standard reduction potential of copper is positive thus, it will undergo a reduction reaction and magnesium will undergo an oxidation reaction.

The half-reaction follows:

Oxidation half-reaction:
`Mg(s)\rightarrow Mg^(2+)(aq)+2e^-`

Reduction half-reaction:
`Cu^(2+)(aq)+2e^-\rightarrow Cu(s)`

Overall cell-reaction:
`Mg(s)+Cu^(3+)(aq)\rightarrow Mg^(2+)(aq)+Cu(s)`

As it can be seen from the reaction that copper is forming as a pure metal in the product thus, it will be deposited and magnesium will be dissolved forming an aqueous solution.

Hence, the correct option is B.