Consider the following four electrodes :
$$\eqalign{
& P = C{u^{2 + }}\left( {0.0001\,M} \right)/Cu\left( s \right) \cr
& Q = C{u^{2 + }}\left( {0.1\,M} \right)/Cu\left( s \right) \cr
& R = C{u^{2 + }}\left( {0.01\,M} \right)/Cu\left( s \right) \cr
& S = C{u^{2 + }}\left( {0.001\,M} \right)/Cu\left( s \right) \cr} $$
If the standard reduction potential of $$C{u^{2 + }}/Cu$$ is $$ + 0.34\,V,$$ the reduction potentials in volts of the above electrodes follow the order.
A.
$$P > S > R > Q$$
B.
$$S > R > Q > P$$
C.
$$R > S > Q > P$$
D.
$$Q > R > S > P$$
Answer :
$$Q > R > S > P$$
Solution :
$${E_{red}} = E_{red}^ \circ + \frac{{0.591}}{n}{\text{log}}\left[ {{M^{n + }}} \right]$$
Lower the concentration of $${{M^{n + }},}$$ lower is the reduction potential.
Hence order of reduction potential is : $$Q > R > S > P$$
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 :