On passing current through two cells, connected in series containing solution of $$AgN{O_3}$$ and $$CuS{O_4},0.18\,g$$ of $$Ag$$ is deposited. The amount of the $$Cu$$ deposited is :
A.
0.529$$\,g$$
B.
10.623$$\,g$$
C.
0.0529$$\,g$$
D.
1.2708$$\,g$$
Answer :
0.0529$$\,g$$
Solution :
$$\eqalign{
& {\text{Using Faraday's second law of electrolysis,}} \cr
& \frac{{{\text{Weight of }}Cu{\text{ deposited}}}}{{{\text{Weight of }}Ag{\text{ deposited}}}} = \frac{{Equ.\,wt.\,{\text{of}}\,Cu}}{{Equ.\,wt.\,{\text{of}}\,Ag}} \cr
& \Rightarrow \frac{{{w_{Cu}}}}{{0.18}} = \frac{{63.5}}{2} \times \frac{1}{{108}} \cr
& \Rightarrow {w_{Cu}} = \frac{{63.5 \times 18}}{{2 \times 108 \times 100}} \cr
& = 0.0529\,g. \cr} $$
Releted MCQ Question on Physical Chemistry >> Electrochemistry
Releted Question 1
The standard reduction potentials at $$298 K$$ for the following half reactions are given against each
$$\eqalign{
& Z{n^{2 + }}\left( {aq} \right) + 2e \rightleftharpoons Zn\left( s \right)\,\,\,\,\,\,\,\,\, - 0.762 \cr
& C{r^{3 + }}\left( {aq} \right) + 2e \rightleftharpoons Cr\left( s \right)\,\,\,\,\,\,\,\,\, - 0.740 \cr
& 2{H^ + }\left( {aq} \right) + 2e \rightleftharpoons {H_2}\left( g \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.000 \cr
& F{e^{3 + }}\left( {aq} \right) + 2e \rightleftharpoons F{e^{2 + }}\left( {aq} \right)\,\,\,\,\,\,\,\,0.770 \cr} $$
which is the strongest reducing agent ?
A solution containing one mole per litre of each $$Cu{\left( {N{O_3}} \right)_2};AgN{O_3};H{g_2}{\left( {N{O_3}} \right)_2};$$ is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are :
$$\eqalign{
& Ag/A{g^ + } = + 0.80,\,\,2Hg/H{g_2}^{ + + } = + 0.79 \cr
& Cu/C{u^{ + + }} = + 0.34,\,Mg/M{g^{ + + }} = - 2.37 \cr} $$
With increasing voltage, the sequence of deposition of metals on the cathode will be :