Question

In a certain electrolysis experiment 0.561 g of Zn is deposited in one cell containing $ZnS{{O}_{4}}$ solution. Calculate the mass of Cu deposited in another cell containing $CuS{{O}_{4}}$ solution in series with $ZnS{{O}_{4}}$ cell.
(Molar masses of Zn and Cu are $65.4gmo{{l}^{-1}}$ and $63.5gmo{{l}^{-1}}$ respectively (0.5447g).

Answer 1

Hint: The equation used here is, $\dfrac{{{W}_{Zn}}}{{{E}_{Zn}}}=\dfrac{{{W}_{Cu}}}{{{E}_{Cu}}}$
Here ${{W}_{Zn}}$ and ${{W}_{Cu}}$ are the weight of Zn and Cu deposited in the cell and ${{E}_{Zn}}$ and ${{E}_{Cu}}$ are the equivalent weight of the Zn and Cu respectively.

Complete step by step answer:
- So in the question it is given that a electrolysis experiment of Zn and Cu is been carried out in two separate cells with their respective solution $ZnS{{O}_{4}}$ and $CuS{{O}_{4}}$. We have the value of the Zn deposited during the electrolysis and we have to find the amount of Cu deposited in series with the $ZnS{{O}_{4}}$ cell.
Let’s write the equation for the electrolysis reaction.
- The reaction of Zn deposition in the cell by the electrolysis of $ZnS{{O}_{4}}$ can be represented as:
$Zn\to Z{{n}^{2+}}+2{{e}^{-}}$
The reaction of Cu deposition in the cell by the electrolysis of $CuS{{O}_{4}}$ can be represented as:
$Cu\to C{{u}^{2+}}+2{{e}^{-}}$
Now take and write the given data in the question.
Weight of the Zn deposited = 0.561 g
Molecular mass of Zn (${{M}_{Zn}}$) = 65.4 g
Molecular mass of Cu (${{M}_{Cu}}$) = 63.5 g
Let’s substitute these values in the equation, $\dfrac{{{W}_{Zn}}}{{{E}_{Zn}}}=\dfrac{{{W}_{Cu}}}{{{E}_{Cu}}}$
- Equivalence mass is the mass equal to the atomic mass divided by the valency of the compound.
$Equivalence\,mass = \dfrac{Atomic\,mass}{Valency\,of\,the\,compound}$
$E=\dfrac{M}{n}$
Here valency (n) of the compound is 2.
$\dfrac{(0.561)}{\left( \dfrac{65.4}{2} \right)}=\dfrac{{{W}_{Cu}}}{\left( \dfrac{63.5}{2} \right)}$
$2{{W}_{Cu}} = 0.0171\times 63.5 = 1.08585$
${{W}_{Cu}} = 0.542$
So the amount of Cu deposited is = 0.542g.

Note: Alternative method:
The equation for the electrolysis reaction of Zn in $ZnS{{O}_{4}}$is,
$Zn\to Z{{n}^{2+}}+2{{e}^{-}}$
The equation for the electrolysis reaction of Cu in $CuS{{O}_{4}}$ is,
$Cu\to C{{u}^{2+}}+2{{e}^{-}}$
We use the equation,$Z=\dfrac{M}{nF}$
65.4g of Zn is deposited =2F charge
0.561g of Zn is deposited by = $\dfrac{2F\times 0.561}{65.4}$ = 0.017F charge
63.5g of Cu is deposited =2Fcharge
Charge deposited by Cu = $\dfrac{63.5\times 0.017F}{2F}$ = 0.5397 = 0.54g