Question

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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.

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.