Question

Calculate the standard free-energy change at 25 ∘C for the following reaction: Mg2+(aq) + Zn(s) → Mg(s) + Zn2+(aq) Express your answer to three significant figures and in units of kJ/mol.

Consider constructing a voltaic cell with one compartment containing a Zn(s) electrode immersed in a Zn²⁺ aqueous solution and the other compartment containing an Al(s) electrode immersed in an Al³⁺ aqueous solution. What is the spontaneous reaction in this cell?

Group of answer choices

Zn + Al³⁺ → Al + Zn²⁺

Al + Zn²⁺ → Zn + Al³⁺

3 Zn + 2 Al³⁺ → 2 Al + 3 Zn²⁺

2 Al + 3 Zn²⁺ → 3 Zn + 2 Al³⁺

Nickel and iron electrodes are used to build a voltaic cell. Based on the standard reduction potentials of Ni²⁺ and Fe³⁺, what is the shorthand notation for this voltaic cell?

Group of answer choices

Ni2+(aq)|Ni(s)||Fe(s)|Fe3+(aq)

Fe3+(aq)|Fe(s)||Ni(s)|Ni2+(aq)

Ni(s)|Ni2+(aq)||Fe3+(aq)|Fe(s)

Fe(s)|Fe3+(aq)||Ni2+(aq)|Ni(s)

Answer 1

For the reaction:

$Mg_{(aq)}^{2+}+Zn_{(s)}\rightarrow Mg_{(s)}+Zn_{(aq)}^{2+}$

The standard change in potential is:

$\Delta E^{o}=E^{o}_{reduction}-E^{o}_{oxidation}$

The standard potentials are:

E°_Zn/Zn²⁺= 0.76 V

E°_Mg/Mg²⁺ = -2.37 V

$\Delta E^{o}=E^{o}_{reduction}-E^{o}_{oxidation}=-2.37V-0.76V=-3.13V$

And to calculate the change in standard free energy, we can use:

$\Delta G^{o}=-nF\Delta E^{o}=-2\cdot 96500\frac{C}{mol}\cdot (-3.13V)=604090\frac{J}{mol}$

Where n is the number of electrons involved in the reaction and F is Faraday's constant.

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The first two choices can be discarded, since they are not balanced in terms of electrical charges, so they can't even be considered reactions.

From the third and four, we can find out which is the spontaneous if we look for the one with a positive delta E°. A positive delta E° is equivalent to a negative delta G°, due to:

$\Delta G^{o}=-nF\Delta E^{o}$

Since

$\Delta E^{o}=E^{o}_{reduction}-E^{o}_{oxidation}$

The half-reaction reaction with the most positive reduction potential will be the one that occurs as a reduction. For the half-reactions, we have:

E°_Zn/Zn²⁺= 0.76 V

and

E°_Al/Al³⁺= -1.66 V

This means that zinc will be reduced and aluminum will be oxidized in the spontaneous process, which will be:

$3Zn_{(aq)}^{2+}+2Al_{(s)}\rightarrow 3Zn_{(s)}+2Al_{(aq)}^{3+}$

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The third question is similar to the second, since it is based on determining which will be the reduction and which the oxidation half-reaction.

We have:

E°_Ni/Ni²⁺= -0.25 V

and

E°_Fe/Fe³⁺= -0.04 V

The most positive is the half-reaction of iron, so this will be the reduction, which is the reaction that is expressed on the right in cell notation. The overall reaction would be:

$2Fe_{(aq)}^{3+}+3Ni_{(s)}\rightarrow 2Fe_{(s)}+3Ni_{(aq)}^{2+}$

And the cell would be noted:

Ni(s)|Ni(2+, aq)||Fe(3+,aq)|Fe(s)