Electrode potentials $$\left( {{E^ \circ }} \right)$$ are given below :
$$\eqalign{
& \frac{{C{u^ + }}}{{Cu}} = + 0.52\,V, \cr
& \frac{{F{e^{3 + }}}}{{F{e^{2 + }}}} = + 0.77\,V, \cr
& \frac{1}{2}\frac{{{I_2}\left( s \right)}}{{{I^ - }}} = + 0.54\,V, \cr
& \frac{{A{g^ + }}}{{Ag}} = + 0.88V. \cr} $$
Based on the above potentials, strongest oxidizing agent will be :
A.
$$C{u^ + }$$
B.
$$F{e^{3 + }}$$
C.
$$A{g^ + }$$
D.
$${I_2}$$
Answer :
$$A{g^ + }$$
Solution :
Higher the value of reduction potential stronger will be the oxidising hence based
on the given values $$A{g^ + }$$ will be strongest oxidizing agent.
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 :