A dilute aqueous solution of $$N{a_2}S{O_4}$$ is electrolyzed using
platinum electrodes. The products at the anode and cathode are :
A.
$${O_2},{H_2}$$
B.
$${S_2}O_8^{2 - },Na$$
C.
$${O_2},Na$$
D.
$${S_2}O_8^{2 - },{H_2}$$
Answer :
$${O_2},{H_2}$$
Solution :
$${H_2}O$$ is more readily reduced at cathode than $$N{a^ + }.$$ It is also more readily oxidized at anode than $$SO_4^{2 - }.$$ Hence, the electrode reactions are
$$\eqalign{
& 2{H_2}O + 2{e^ - } \to {H_2} \uparrow + 2O{H^ - }\left[ {{\text{at cathode}}} \right] \cr
& {H_2}O \to \frac{1}{2}{O_2} \uparrow + 2{H^ + } + 2{e^ - }\left[ {{\text{at anode}}} \right] \cr} $$
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 :