Question

For the following voltaic cell, write the half-reactions, designating which is oxidation and which is reduction. Write the cell reaction and calculate the voltage of the cell made from standard electrodes. The cell is \(\mathrm{Co} ; \mathrm{Co}^{+2} \mid \mathrm{Ni}^{+2} ; \mathrm{Ni}\).

Short answer

The oxidation half-reaction is \(\mathrm{Co} \rightarrow \mathrm{Co}^{2+} + 2\mathrm{e}^-\) and the reduction half-reaction is \(\mathrm{Ni}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Ni}\). The balanced cell reaction is \(\mathrm{Co} + \mathrm{Ni}^{2+} \rightarrow \mathrm{Co}^{2+} + \mathrm{Ni}\). Using standard reduction potentials, we calculate the cell voltage as \(E_\mathrm{cell}^\circ = -0.51\;V\).

Step-by-step solution

1. Identify the Oxidation and Reduction Half-Reactions

: We have two half-reactions in the given voltaic cell, involving Co and Ni ions: - \(\mathrm{Co} \rightarrow \mathrm{Co}^{2+} + 2\mathrm{e}^-\) - \(\mathrm{Ni}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Ni}\) In the first half-reaction, Co is losing electrons, which means this is an oxidation half-reaction. In the second half-reaction, Ni ions are gaining electrons, so it is a reduction half-reaction.

2. Write the Balanced Cell Reaction

: Now that we have identified the oxidation and reduction half-reactions, we can write the balanced cell reaction by combining them: - Oxidation half-reaction: \(\mathrm{Co} \rightarrow \mathrm{Co}^{2+} + 2\mathrm{e}^-\) - Reduction half-reaction: \(\mathrm{Ni}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Ni}\) Combined cell reaction: \(\mathrm{Co} + \mathrm{Ni}^{2+} \rightarrow \mathrm{Co}^{2+} + \mathrm{Ni}\)

3. Calculate the Cell Voltage using Standard Reduction Potentials

: To calculate the voltage of the cell, we need to use the standard reduction potentials for each of the half-reactions. The standard reduction potentials are: - \(\mathrm{Co}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Co} \; E^\circ = -0.28\;V\) - \(\mathrm{Ni}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Ni} \; E^\circ = -0.23\;V\) As they are reduction potentials, we must reverse the sign of the oxidation half-reaction (Co): - \(\mathrm{Co} \rightarrow \mathrm{Co}^{2+} + 2\mathrm{e}^- \; E^\circ = 0.28\;V\) Now, we can calculate the cell voltage by adding the standard reduction potentials: - \(E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ - E_\mathrm{ox}^\circ = (-0.23\;V) - (0.28\;V) = -0.51\;V\) Since the voltage is negative, this means the reaction is not spontaneous in the direction written. Instead, the spontaneous reaction would be the reverse of the given cell reaction. However, we have successfully determined the cell voltage for the given voltaic cell reaction as -0.51 V.

Key concepts

Oxidation Half-Reaction

In a voltaic cell, oxidation occurs when a substance loses electrons. For the Cobalt (Co) involved in our voltaic cell, the half-reaction is:

• \(\mathrm{Co} \rightarrow \mathrm{Co}^{2+} + 2\mathrm{e}^-\)

This reaction shows Cobalt transforming into Cobalt ions (\(\mathrm{Co}^{2+}\)), emitting two electrons in the process.
This release of electrons is a typical representation of oxidation. Remember, when a species' oxidation state increases, it signifies oxidation.
Retracing our steps: Cobalt started as a element and lost two electrons, proving it acts as the source of electrons! By identifying this, we know this reaction must be happening at the anode of the voltaic cell.

Reduction Half-Reaction

Reduction happens when a substance gains electrons. In our cell, Nickel ions (\(\mathrm{Ni}^{2+}\)) undergo reduction:

• \(\mathrm{Ni}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Ni}\)

Here, Nickel ions gain two electrons to form solid Nickel. This gain is key: reduction is all about a species accepting electrons.
When an ion collects electrons, its oxidation state diminishes, confirming reduction.
The reduction process takes place at the cathode of the cell. Thus, we see Nickel ions bridging the loss of electrons from the oxidation process!

Standard Reduction Potential

Standard Reduction Potentials (\(E^\circ\)) are crucial for predicting and calculating cell voltage.
These potentials measure the tendency of a substance to gain electrons and be reduced. For Cobalt and Nickel:

• \(\mathrm{Co}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Co} \; E^\circ = -0.28\;V\)
• \(\mathrm{Ni}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Ni} \; E^\circ = -0.23\;V\)

The standard table usually lists these as reductions.
To tackle the oxidation half-reaction, reverse the sign.
This is necessary since electrons emitted during oxidation need to be represented positively!
So, for the oxidation of Cobalt, the potential becomes:

• \(\mathrm{Co} \rightarrow \mathrm{Co}^{2+} + 2\mathrm{e}^- \; E^\circ = 0.28\;V\)

This adjustment helps in correctly determining the overall cell voltage.

Cell Voltage Calculation

Cell Voltage calculation showcases the potential difference between the cathode and anode.
It determines if a reaction is spontaneous. To find it:

• Calculate by combining the standard reduction potential of the reduction half-reaction and the reversed oxidation potential.

For our cell, the equation looks like this:

• \[E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ - E_\mathrm{ox}^\circ\]

Substitute values:

• \(E_\mathrm{cell}^\circ = (-0.23\;V) - (0.28\;V) = -0.51\;V\)

A negative voltage indicates non-spontaneity for the cell written as is.
This means, under standard conditions, the given reaction won't proceed. Yet, this approach lays down the groundwork for understanding how cells operate and predict their behavior!