Given
$$\eqalign{
& F{e^{3 + }}\left( {aq} \right) + {e^ - } \to F{e^{2 + }}\left( {aq} \right);{E^ \circ } = + 0.77\,V \cr
& A{l^{3 + }}\left( {aq} \right) + 3{e^ - } \to Al\left( s \right);{E^ \circ } = - 1.66\,V \cr
& B{r_2}\left( {aq} \right) + 2{e^ - } \to 2B{r^ - };{E^ \circ } = + 1.09\,V \cr} $$
Considering the electrode potentials, which of the following represents the correct order of reducing power ?
A.
$$F{e^{2 + }} < Al < B{r^ - }$$
B.
$$B{r^ - } < F{e^{2 + }} < Al$$
C.
$$Al < B{r^ - } < F{e^{2 + }}$$
D.
$$Al < F{e^{2 + }} < B{r^ - }$$
Answer :
$$Al < F{e^{2 + }} < B{r^ - }$$
Solution :
Reducing character decreases down the series. Hence the correct order is
$$Al < F{e^{2 + }} < B{r^ - }$$
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 :