Chapter U5.19, Problem 3E

Solution

(a)

Interpretation:The half-reactions for each beaker needs to be determined.

Concept Introduction : A electrochemical reaction is marked by the presence of two half cells. The overall voltage is measured as the difference between the potentials of the half cells in an electrochemical cell.

The half reactions are,

  $\begin{array}{l}\text{Ca}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\text{+}{\text{2e}}^{-}\\ \text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{3e}}^{-}\end{array}$

Given that, one beaker has a calcium metal electrode and a solution of aqueous calcium ions, ${\text{Ca}}^{\text{2+}}\left(\text{aq}\right)$ and the other beaker contains iron metal electrode and a solution of aqueous iron ions, ${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)$

Therefore, the half reactions are,

  $\begin{array}{l}\text{Ca}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\text{+}{\text{2e}}^{-}\\ \text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{3e}}^{-}\end{array}$

(b)

Interpretation:The metal being oxidized and reduced needs to be determined.

Concept Introduction : A electrochemical reaction is marked by the presence of two half cells. The overall voltage is measured as the difference between the potentials of the half cells in an electrochemical cell.

Calcium is oxidized and iron is reduced.

The activity series of metal in the order of increasing reactivity is,

Gold < Silver < Mercury < Bismuth < Antimony < Hydrogen < Lead < Tin < Nickel < Cobalt < Cadmium < Iron < Chromium < Zinc < Manganese < Aluminium < Calcium < Strontium < Barium < Lithium < Sodium < Potassium

From the activity series it can be concluded that, calcium has is more active than iron. Hence, calcium is oxidized and iron is reduced.

(c)

Interpretation:A sketch of the electrochemical cell needs to drawn and the reaction that occur in each beaker needs to be shown.

Concept Introduction : A electrochemical reaction is marked by the presence of two half cells. The overall voltage is measured as the difference between the potentials of the half cells in an electrochemical cell.

A sketch of the electrochemical cell showing the reaction that occur in each beaker is given below:

  

  

Fig:A sketch of the electrochemical cell showing the reaction that occur in each beaker.

The half reactions on both the beakers are,

  $\begin{array}{l}\text{Ca}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\text{+}{\text{2e}}^{-}\\ \text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{3e}}^{-}\end{array}$

(d)

Interpretation:The net ionic equation for the reaction that takes place to generate electricity needs to be determined.

Concept Introduction:A electrochemical reaction is marked by the presence of two half cells. The overall voltage is measured as the difference between the potentials of the half cells in an electrochemical cell.

The net ionic equation is,

  $\text{3Ca}\left(\text{s}\right)+2{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\to 2\text{Fe}\left(\text{s}\right)+3{\text{Ca}}^{\text{2+}}\left(\text{aq}\right)$

Given that, one beaker has a calcium metal electrode and a solution of aqueous calcium ions, ${\text{Ca}}^{\text{2+}}\left(\text{aq}\right)$ and the other beaker contains iron metal electrode and a solution of aqueous iron ions, ${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)$

Therefore, the half reactions on the beakers are,

  $\begin{array}{l}\text{Ca}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\text{+}{\text{2e}}^{-}\\ \text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{3e}}^{-}\end{array}$

And the net ionic equation is obtained as:

  $\begin{array}{l}\left[\text{Ca}\left(\text{s}\right)\to {\text{Ca}}^{\text{2+}}\text{+}{\text{2e}}^{-}\right]\times 3\\ \left[\text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{3e}}^{-}\right]\times 2\end{array}$

So,

  $\begin{array}{l}\text{3Ca}\left(\text{s}\right)\to 3{\text{Ca}}^{\text{2+}}\text{+}6{\text{e}}^{-}\\ \underset{\_}{2{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}6{\text{e}}^{-}\to \text{2Fe}\left(\text{s}\right)}\\ \text{3Ca}\left(\text{s}\right)+2{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\to 2\text{Fe}\left(\text{s}\right)+3{\text{Ca}}^{\text{2+}}\left(\text{aq}\right)\end{array}$