- In order to calculate the mass of the products, we have to look at the steichiometry of the chemical reaction.
- That is, we have to compare the number of electrons to the number of product atoms or molecules and use this to calculate the amount of the products.
- When we have this, we can easily calculate the mass via the molar mass of the product.
(a) Cathode
\(C{a^{2 + }}(l) + 2{e^ - } \to Ca(s)\)
Here we see that
\(n\left( {{e^ - }} \right):n(Ca) = 2:1\)
So
\(n(Ca) = \frac{1}{2} \cdot n\left( {{e^ - }} \right)\)
We can calculate the \(n\left( {{e^ - }} \right)\) using the formula
\(Q = N\left( {{e^ - }} \right) \cdot e\)
Where\(e\) is the electric charge of a single electron.
This is a constant and it is\(1.6 \cdot {10^{ - 19}}C\).
\(N\left( {{e^ - }} \right) = \frac{Q}{e}\)
\(N\left( {{e^ - }} \right) = 2.08 \cdot {10^{24}}\)
\(n\left( {{e^ - }} \right) = \frac{{N\left( {{e^ - }} \right)}}{{{N_A}}}\)
(\({N_A}\)Is the Avogadro constant and it is \(6.022 \cdot {10^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\) )
\(n\left( {{e^ - }} \right) = 3.45\;{\rm{mol}}\)
\(n(Ca) = \frac{1}{2} \cdot 3.45\;{\rm{mol}}\)
\(n(Ca) = 1.725\;{\rm{mol}}\)
\(M(Ca) = Ar(Ca) \cdot \frac{g}{{mol}}\)
(\(Ar\) is relative atomic mass that can be found in the Periodic table)
\(M(Ca) = 40.078\frac{g}{{mol}}\)
\(m(Ca) = n(Ca) \cdot M(Ca)\)
\({\rm{ }}m(Ca) = 69.13g\)
A) anode:
\(2C{l^ - }(l) \to C{l_2}(g) + 2{e^ - }\)
Since we already calculated the amount of electrons in the previous task, we do not need to do so here. From the equation on the anode, we can calculate the ratio between amount of the electrons and the chlorine:
\(n\left( {{e^ - }} \right):n\left( {C{l_2}} \right) = 2:1\)
So,
\(n\left( {C{l_2}} \right) = \frac{1}{2} \cdot m\left( {{e^ - }} \right)\)
\(m\left( {C{l_2}} \right) = 1.725mol\)
To calculate the mass we have to first calculate molar mass of the chlorine:
\(M\left( {C{l_2}} \right) = 2 \cdot Ar(Cl)\frac{g}{{mol}}\)
\(M\left( {C{l_2}} \right) = 70.9\frac{g}{{mol}}\)
So, mass of the chlorine is:
\(m\left( {C{l_2}} \right) = n\left( {C{l_2}} \right) \cdot M\left( {C{l_2}} \right)\)
\(m\left( {C{l_2}} \right) = 122.303g\)
b) Cathode:
\(L{i^ + }(l) + {e^ - } \to Li(s)\)
\(n(Li):n\left( {{e^ - }} \right) = 1:1\)
We already calculated the amount of electrons in the task a.
\(n(Li) = n\left( {{e^ - }} \right) = 3.45\;{\rm{mol}}\)
\(m(Li) = n(Li) \cdot M(Li)\)
\(M(Li) = 6.941\frac{g}{{mol}}\)
\(m(Li) = 23.946g\)
Anode:
\(2{{\rm{H}}^ - }(l) \to {{\rm{H}}_2}(g) + 2{e^ - }\)
\(n\left( {{H_2}} \right):n\left( {{e^ - }} \right) = 1:2\)
\(n\left( {{H_2}} \right) = \frac{1}{2} \cdot n\left( {{e^ - }} \right)\)
\(n\left( {{H_2}} \right) = 1.725\;{\rm{mol}}\)
\(M\left( {{H_2}} \right) = 2.106\frac{g}{{mol}}{\rm{ }}\)
\(m\left( {{H_2}} \right) = 3.478g\)
c) Cathode:
\(A{l^{3 + }}(l) + 3{e^ - } \to Al(s)\)
\(n(Al):n\left( {{e^ - }} \right) = 1:3\)
\(n(Al) = \frac{1}{3} \cdot n\left( {{e^ - }} \right)\)
\(n(Al) = 1.15\;{\rm{mol}}\)
\(M(Al) = 26.98\frac{g}{{mol}}\)
\(m(Al) = 31.027g\)
Anode:
\(2C{l^ - }(l) \to C{l_2}(g) + 2{e^ - }\)
\(n\left( {C{l_2}} \right):n\left( {{e^ - }} \right) = 1:2\)
\(n\left( {C{l_2}} \right) = \frac{1}{2} \cdot n\left( {{e^ - }} \right)\)
\(n\left( {C{l_2}} \right) = 1.725\;{\rm{mol}}\)
\(M\left( {C{l_2}} \right) = 70.9\frac{g}{{mol}}\)
\(m\left( {C{l_2}} \right) = 122.303g\)
d) Cathode:
\(C{r^{3 + }}(l) + 3{e^ - } \to Cr(s)\)
\(n(Cr):n\left( {{e^ - }} \right) = 1:3\)
\(n(Cr) = \frac{1}{3} \cdot n\left( {{e^ - }} \right)\)
\(n(Cr) = 1.15\;{\rm{mol}}\)
\(M(Cr) = 51.996\frac{g}{{mol}}\)
\(m(Cr) = 59.796g\)
Anode:
\(n\left( {B{r_2}} \right):n\left( {{e^ - }} \right) = 1:2\)
\(n\left( {B{r_2}} \right) = \frac{1}{2} \cdot n\left( {{e^ - }} \right)\)
\(n\left( {B{r_2}} \right) = 1.725\;{\rm{mol}}\)
\(M\left( {B{r_2}} \right) = 159.808\frac{g}{{mol}}\)
\(m\left( {B{r_2}} \right) = 275.669g\)
The mass of each product is
(a) \(\begin{aligned}{{}{}}{m(Ca) = 69.13g}\\{m\left( {C{l_2}} \right) = 122.303g}\end{aligned}\)
(b) \(\begin{aligned}{{}{}}{m(Li) = 23.946g}\\{m\left( {{H_2}} \right) = 3.478g}\end{aligned}\)
(c) \(\begin{aligned}{{}{}}{m(Al) = 31.027g}\\{m\left( {C{l_2}} \right) = 122.303g}\end{aligned}\)
(d) \(\begin{aligned}{{}{}}{m(Cr) = 59.796g}\\{m\left( {B{r_2}} \right) = 275.669g}\end{aligned}\)
