Electrode potential data of few cells is given below. Based on the data, arrange the ions in increasing order of their reducing power.
$$\eqalign{
& Fe_{\left( {aq} \right)}^{3 + } + {e^ - } \to Fe_{\left( {aq} \right)}^{2 + };{E^ \circ } = + 0.77\,V \cr
& Al_{\left( {aq} \right)}^{3 + } + 3{e^ - } \to A{l_{\left( s \right)}};{E^ \circ } = - 1.66\,V \cr
& B{r_{2\left( {aq} \right)}} + 2{e^ - } \to 2Br_{\left( {aq} \right)}^ - ;{E^ \circ } = + 1.09\,V \cr} $$
A.
$$B{r^ - } < F{e^{2 + }} < Al$$
B.
$$F{e^{2 + }} < Al < B{r^ - }$$
C.
$$Al < B{r^ - } < F{e^{2 + }}$$
D.
$$Al < F{e^{2 + }} < B{r^ - }$$
Answer :
$$B{r^ - } < F{e^{2 + }} < Al$$
Solution :
Lower the reduction potential, more is the reducing power.
$$B{r^ - } < F{e^{2 + }} < Al$$
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 :