Use the data given below and find out the most stable ion in its reduced form.
$$\eqalign{
& E_{\frac{{C{r_2}O_7^{2 - }}}{{C{r^{3 + }}}}}^ \circ = 1.33\,V;E_{\frac{{C{l_2}}}{{C{l^ - }}}}^ \circ = 1.36\,V \cr
& E_{\frac{{MnO_4^ - }}{{M{n^{2 + }}}}}^ \circ = 1.51\,V;E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ = - 0.74\,V \cr} $$
A.
$$C{l^ - }$$
B.
$$C{r^{3 + }}$$
C.
$$Cr$$
D.
$$M{n^{2 + }}$$
Answer :
$$M{n^{2 + }}$$
Solution :
$$E_{\frac{{MnO_4^ - }}{{M{n^{2 + }}}}}^ \circ $$ has highest reduction potential hence, $$M{n^{2 + }}$$ is the most stable reduced species.
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 :