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Electrochemistry — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry4 Marks
Question
A constant electric current flows for 4 hours through two electrolytic cells connected in series. One contains $AgNO _3$ solution and second contains $CuCl _2$ solution. During this time, 4 grams of $Ag$ are deposited in the first cell.
(a) How many grams of $Cu$ are deposited in the second cell?
(b) What is the current flowing in amperes? (Atomic mass: $Cu =63.5 gmol ^{-1} ; Ag =107.9 gmol ^{-1}$ )
✓

Answer

Given : Mass of $Ag$ deposited $=4 g$
Molar mass of $Cu =63.5 g mol ^{-1}$
Molar mass of $Ag =107.9 g mol ^{-1}$
Time $= t =4 hrs =4 \times 60 \times 60=14400 s$
Mass of $Cu$ deposited $= W _{ Cu }=$ ?
Current $=1=$ ?
(a) Number of moles of $Ag$ deposited
$\begin{aligned}
& =\frac{\text { Mass of } Ag }{\text { Molar mass of } Ag }=\frac{4}{107.9} \\
& =0.03707 \text { mol of } Ag
\end{aligned}$
Reactions of electrolysis :
(i) $Ag ^{+}+ e ^{-} \rightarrow Ag$ (Half reaction in $AgNO _3$ cell)
(ii) $Cu ^{2+}+2 e ^{-} \rightarrow Cu$ (Half reaction in $CuCl _2$ cell)
Mole ratio of $Ag$
$\begin{aligned}
& =\frac{\text { moles of } Ag \text { produced }}{\text { number of moles of electrons }} \\
& =\frac{1}{1}=1
\end{aligned}$
Mole ratio of $Cu =\frac{1}{2}=0.5$
$\frac{\text { moles of } Cu \text { produced }}{\text { moles of } Ag \text { produced }}=\frac{\text { mole ratio of } Cu }{\text { mole ratio of } Ag }$
$\therefore$ moles of $Cu$ produced
$=\frac{\text { mole ratio of } Cu }{\text { mole ratio of } Ag } \times \text { moles of } Ag$
$=\frac{0.5}{1} \times 0.03707=0.01854 mol$
$\therefore$ Mass of Cu produced $=0.01854 \times 63.5=1.177 g$


(b) From the reaction,
$\because 1 mol Ag { }^{+}$requires $1 mol$ electrons
$\therefore 0.03707 mol Ag$ will require 0.03707 mol electrons
$\because 1$ mol electrons $=1$ Faraday
$\therefore 0.03707$ mol electrons $=0.03707$ Faraday
$\because 1$ Faraday $=96500 C$
$\therefore 0.03707$ Faraday
$=0.03707 \times 96500=3577 C$
$\therefore$ Quantity of electricity $= Q =3577 C$.
$\begin{aligned}
& Q = I \times t \\
& \therefore I =\frac{ Q }{t}=\frac{3577}{14400}=0.25 A
\end{aligned}$
(a) Mass of $Cu$ deposited $=1.177 g$
(b) Current passed $=0.25 A$

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