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Electrochemistry — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry4 Marks
Question
Explain quantitative aspects of electrolysis.
✓

Answer

(1) Calculation of quantity of electricity : If an electric current of strength I $A$ is passed through the cell for $t$ seconds, then quantity of electricity $( Q$ ) obtained is given by,
$Q = I \times t C$ (Coulomb)

(2) Calculation of moles of electrons passed: The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C, then moles of electrons passed $=$ $\frac{Q( C )}{F\left( C / mole ^{-}\right)}$

(3) Calculation of moles of product formed : Consider one mole of ions, $M _{( aq )}^{n^{+}}$which will require moles of electrons for reduction.
$\begin{aligned}
& M _{(\text {aq) }}^{n^{+}}+\text {ne }^{-} \rightarrow M \text { (Reduction half reaction) } \\
& \therefore \text { Mole ratio } \\
& =\frac{\text { moles of product formed in the half reaction }}{\text { moles of electrons required in the half reaction }} \\
& \therefore \text { Moles of product formed } \\
& =\text { moles of electrons passed } \times \text { mole ratio } \\
& =\frac{Q}{96500} \times \text { mole ratio } \\
& =\frac{I \times t}{96500} \times \text { mole ratio }
\end{aligned}$

(4) Calculation of mass of product : Mass, $W$ of product formed is given by,
$W =$ moles of product $\times$ molar mass of product $( M )$
$\begin{aligned}
& =\frac{Q}{96500} \times \text { mole ratio } \times M \\
& =\frac{I \times t}{96500} \times \text { mole ratio } \times M \text {}
\end{aligned}$
When two electrolytic cells containing different electrolytes are connected in series so that same quantity of electricity is passed through them, then the masses $W _1$ and $W _2$ of products produced are given by,
$\begin{aligned}
& W_1=\frac{Q}{96500} \times(\text { mole ratio })_1 \times M_1 \\
& W_2=\frac{Q}{96500} \times(\text { mole ratio })_2 \times M_2 \\
& \therefore \frac{W_2}{W_1}=\frac{(\text { mole ratio })_2 \times M_2}{(\text { mole ratio })_1 \times M_1} \text {} \\
& \therefore \frac{(\text { mole ratio })_2}{(\text { mole ratio })_1}=\frac{W_2}{W_1} \times \frac{M_1}{M_2}=\frac{W_2 / M_2}{W_1 / M_1}=\frac{n_2}{n_1}
\end{aligned}$

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