An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a \({\bf{Zn}}{\left( {{\bf{N}}{{\bf{O}}_3}} \right)_2}\)solution. When a current of \({\bf{2}}.{\bf{599}}{\rm{ }}{\bf{A}}\)was used, it took exactly \({\bf{1}}\) an hour to deposit a \({\bf{0}}.{\bf{01123}} - {\bf{mm}}\) layer of zinc on the part. What was the total surface area of the part? The density of zinc is \({\bf{7}}.{\bf{140}}{\rm{ }}{\bf{g}}/{\bf{c}}{{\bf{m}}^3}.\) Assumed the efficiency is \({\bf{100}}\% \).

Electrochemistry is the study of the generation of electricity from the energy released during spontaneous chemical reactions, as well as the application of electrical energy to non-spontaneous chemical transformations.

Since zinc was deposited, it means that elementary zinc has formed. And since elementary zinc has formed, it means that the \(Z{n^{2 + }}\)ions have been reduced. Knowing this, we can write the cathode half-reaction (reduction always occurs at the cathode, never at the anode):

\(Z{n^{2 + }}(aq) + 2{e^ - } \to Zn(s)\)Knowing the electrical current and the time, we can calculate the total electric charge using the formula

\(Q = I \cdot t\)

\(Q = 9365.4C\)

Now that we know the electric charge, we can use it to calculate the amount of zinc formed by calculating the amount of electrons first.

Now that we know the electric charge, we can use it to calculate the amount of zinc formed by calculating the amount of electrons first.

\(Q = N\left( {{e^ - }} \right) \cdot e\)

(\(e\)is the charge of one single electron,\(1.6 \cdot {10^{ - 19}}{\rm{C}}\))

\(N\left( {{e^ - }} \right) = \frac{Q}{e}\)

\(N\left( {{e^ - }} \right) = 5.9 \cdot {10^{22}}\)

\(n\left( {{e^ - }} \right) = \frac{{N\left( {{e^ - }} \right)}}{{{N_A}}}\)

\(\left( {{N_A}} \right.\)is the Avogadro constant, \(\left. {6.022 \cdot {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)\)

\(m\left( {{e^ - }} \right) = 0.1mol\)

From the reaction equation, we know the ratio of the amount of electrons to the number of zinc atoms and we can therefore calculate the amount of zinc formed during this electrolysis.

\(n(Zn):m\left( {{e^ - }} \right) = 1:2\)

\(n(Zn) = \frac{1}{2} \cdot n\left( {{e^ - }} \right)\)

\(n(Zn) = 0.05mol\)

From the amount of zinc formed, we can calculate its mass

\(m(Zn) = n(Zn) \cdot M(Zn)\)

\(M\)is the molar mass of zinc and we look up its value at the periodic table of elements.

\(m(Zn) = 0.05\;{\rm{mol}} \cdot 65.38\frac{g}{{mol}}\)

\({\rm{ }}m(Zn) = 3.27g\)

Now that we know the mass and the density of zinc, we can calculate the volume of zinc formed.

\(V(Zn) = \frac{{m(Zn)}}{{\rho (Zn)}}\)

\(V(Zn) = 0.458\;{\rm{c}}{{\rm{m}}^3}\)

Since we know the volume and the thickness of the zinc layer, and we can assume that the layer is cuboid, we can calculate the surface area of the layer.

\(V = A \cdot d\)

(Here\(A\)is the surface area and \(d\) is the thickness)

\(A = \frac{V}{d}\)

\(A = \frac{{0.458\;{\rm{c}}{{\rm{m}}^3}}}{{0.01123\;{\rm{m}}{{\rm{m}}^3}}}\)

\(A = \frac{{0.458\;{\rm{c}}{{\rm{m}}^3}}}{{0.001123\;{\rm{c}}{{\rm{m}}^2}}}\)

\(A = 407.836\;{\rm{c}}{{\rm{m}}^2}\)

The total surface area of the zinc layer is \(407.836\;{\rm{c}}{{\rm{m}}^2}\).