Question

What is an electrochemical theory of rusting of irons and give two methods of preventing the rusting of iron?
Write the Nernst equation and calculate the emf of following cell at 298 K:
$\text{Ni/N}{{\text{i}}^{2+}}\left( \text{0}\text{.01} \right)\parallel \text{C}{{\text{u}}^{2+}}/\text{Cu}\left( 0.01\text{M} \right)$
Given $\text{E}{}^\circ \left( \text{C}{{\text{u}}^{2+}}/\text{Cu} \right)\text{ = +0}\text{.34V, E}{}^\circ \left( \text{N}{{\text{I}}^{2}}/\text{Ni} \right)\text{ = -0}\text{.22V}$

Answer 1

Hint: The rusting of iron is a slow process in which the iron reacts with oxygen and forms an iron oxide which is usually reddish-brown in colour. We usually study the Nernst equation and the related numerical in electrochemistry.

Complete Step-by-Step Answer:
-According to the electrochemical theory of the rusting of iron, oxidation reaction occurs at anode and reduction reaction occurs at the cathode.
-At the anode, the iron is oxidised to the ferrous ions and the electrons which are released move towards the cathode.
At the anode: $\text{Fe (s) }\to \text{ F}{{\text{e}}^{2+}}\text{ + 2}{{\text{e}}^{-}}$
-At the cathode, the hydrogen ions are taken from the moisture that is present in the atmosphere.
-So, the reaction of water releasing hydrogen ion is:
\[{{\text{H}}_{2}}\text{O }\to \text{ }{{\text{H}}^{+}}\text{ + O}{{\text{H}}^{-}}\]
-At the cathode:
\[{{\text{H}}^{+}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ H}\]
-But the oxygen that is present on the surface of the iron is reduced by hydrogen. So, the reaction becomes $\text{4}{{\text{H}}^{+}}\text{ + }{{\text{O}}_{2}}\text{ +4}{{\text{e}}^{-}}\text{ }\to \text{ 2}{{\text{H}}_{2}}\text{O}$
-Hence, by using the oxidation and reduction half-reaction we can write the overall reaction i.e.
$\text{2Fe + 4}{{\text{H}}^{+}}\text{ + }{{\text{O}}_{2}}\text{ }\to \text{ 2F}{{\text{e}}^{2+}}\text{ 2}{{\text{H}}_{2}}\text{O}$
-This ferrous ion is further oxidized by the oxygen which forms rust:
$\text{4F}{{\text{e}}^{2+}}\text{ + }{{\text{O}}_{2}}\text{ + 4}{{\text{H}}_{2}}\text{O }\to \text{ 2Fe}{{\text{O}}_{3}}\text{ + 8}{{\text{H}}^{+}}$
Two methods to prevent the rusting are:
-If we will paint (water-insoluble) the metal then it reduces the chances of rusting by separating the metal from the water.
-We can also use oil or grease to keep the water away from the iron.
Nernst Equation: The Nernst equation is the relation between the standard cell potential, temperature, cell potential of an electrochemical cell and the reaction quotient.
${{\text{E}}_{\text{Cell}}}\text{ = E}{{{}^\circ }_{\text{Cell}}}\text{ - }\frac{0.0592}{\text{n}}\log \text{Q}$
-Here, $\text{E}{{{}^\circ }_{\text{Cell}}}$ is a standard electrode potential and log Q is the reaction quotient.
-To calculate the emf, we will use the formula of the Nernst equation:
$\begin{align}
  & {{\text{E}}_{Cell}}\text{ = E}{{{}^\circ }_{\text{C}{{\text{u}}^{2+}}/\text{Cu}}}\text{ + E}{{{}^\circ }_{\text{N}{{\text{i}}^{2+}}/\text{Ni}}}\text{ - }\frac{0.059}{2}\text{log}\left( \frac{{{\text{a}}_{\text{N}{{\text{i}}^{2+}}}}}{{{\text{a}}_{\text{C}{{\text{u}}^{2+}}}}} \right) \\
 & {{\text{E}}_{Cell}}\text{ = 0}\text{.34V -}\left( -0.22\text{V} \right)\text{ - }\frac{0.059}{2}\text{ log }\left( \frac{\text{0}\text{.01}}{0.01} \right) \\
 & {{\text{E}}_{\text{Cell}}}\text{ = 0}\text{.66V} \\
\end{align}$

Therefore, the value of emf of the cell is 0.66V.

Note: The value of standard temperature is 298 Kelvin. Increase in the oxidation state means the oxidation whereas a decrease in the oxidation state means the reduction of a molecule. In the given question, nickel undergoes the oxidation reaction whereas copper undergoes the reduction reaction.