Identifying Components of a Voltaic Cell Magnesium reacts with copper(II) sulfate according to the following equation: Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s) (a) Write equations to represent the oxidation and reduction half-reactions. (b) Determine the oxidizing and reducing agents. (c)

Question

Identifying Components of a Voltaic Cell

Magnesium reacts with copper(II) sulfate according to the following equation:

Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)

(a) Write equations to represent the oxidation and reduction half-reactions.

(b) Determine the oxidizing and reducing agents.

(c) The figure represents components of a voltaic cell in which this reaction

takes place. Answer the following questions with the correct component of

a voltaic cell, using the letters A to E as shown. Assume that the oxidation

half-reaction will occur on the left.

In which half-cell will you put the oxidation components?
In which half-cell will you put the reduction components?
What will you use for the anode and cathode?
What electrolyte solutions will you put in each compartment?
Which direction will ions flow from the salt bridge?
Where do electrons flow?

Accepted Answer

(a) Because the sulfate ion, SO₄^2–, is a spectator ion, we can simplify the reaction by writing a net ionic equation:

Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)

Since these are neutral or ionic species, assigning oxidation numbers is straightforward:

[latex]\overset{\overset{0}{\downarrow } }{M}g(s)+\overset{\overset{2+}{\downarrow } }{C}u^{2+}(aq)\longrightarrow \overset{\overset{2+}{\downarrow } }{M}g^{2+}(aq)+\overset{\overset{0}{\downarrow } }{C}u(s)[/latex]

Magnesium’s oxidation number changes from 0 to 2+, so it is the reactant that is oxidized. The oxidation reaction is a loss of two electrons for each magnesium atom:

Mg(s) → Mg²⁺ (aq) + 2e^–

Copper’s oxidation number changes from 2+ to 0, so it is the reactant that’s reduced. The reduction reaction is a gain of two electrons for each Cu²⁺ ion:

Cu²⁺ (aq) + 2e^– →Cu(s)

From this information we can write half-reaction equations that represent the oxidation and reduction processes:

Oxidation Mg(s) → Mg²⁺ (aq) + 2e^–

Reduction Cu²⁺ (aq) + 2e^– →Cu(s)

(b) Magnesium is the reactant that’s oxidized. The reactant that accepts the electrons from magnesium is Cu²⁺. The oxidizing agent is Cu²⁺, which is reduced in the reaction. The reactant that provides the electrons for this reduction is magnesium. Magnesium is the reducing agent.

(c) If we arbitrarily assign the oxidation half-reaction to the compartment on the left, then the answers are as follows:

The oxidation components will be A and B on the left.
The right compartment will contain the reduction components D and E.
Component A will be the anode of magnesium metal. Component E will be the cathode of copper metal.
The solution in the anode compartment, B, should be an electrolyte of magnesium, such as Mg(NO₃)₂. The electrolyte solution in the cathode compartment, D, can be Cu(NO₃)₂ or another salt containing Cu²⁺ ions.
To complete the electric circuit, component C needs to be a salt bridge that allows ions to flow into the two half-cells. Sodium sulfate, Na₂SO₄, is a common ingredient in salt bridges. Sulfate ions, SO₄^2–, will flow into the left compartment to balance the positive charge that builds up from the production of Mg²⁺ ions. Na⁺ ions will flow into the right half-cell to replace the Cu²⁺ ions that react to form neutral Cu atoms.
Electrons will flow from electrode A to electrode E through, in this case, a voltmeter.
The mass of electrode A will decrease because magnesium atoms react to form magnesium ions that flow into the solution. The mass of electrode E will increase because copper ions from the solution react to form copper atoms on the surface of the electrode.

Question 14.4: Identifying Components of a Voltaic Cell Magnesium reacts wi...

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