Question

Using the data given below find out in which option the order of reducing power is correct.
$$\eqalign{ & E_{\frac{{C{r_2}O_7^{2 - }}}{{C{r^{3 + }}}}}^ \circ = 1.33\,V;E_{\frac{{C{l_2}}}{{C{l^ - }}}}^ \circ = 1.36\,V \cr & E_{\frac{{MnO_4^ - }}{{M{n^{2 + }}}}}^ \circ = 1.51\,V;E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ = - 0.74\,V \cr} $$

A. $$C{r^{3 + }} < C{l^ - } < M{n^{2 + }} < Cr$$
B. $$M{n^{2 + }} < C{l^ - } < C{r^{3 + }} < Cr$$  
C. $$C{r^{3 + }} < C{l^ - } < C{r_2}O_7^{2 - } < MnO_4^ - $$
D. $$M{n^{2 + }} < C{r^{3 + }} < C{l^ - } < Cr$$
Answer :   $$M{n^{2 + }} < C{l^ - } < C{r^{3 + }} < Cr$$
Solution :
Lower the reduction potential, higher is the reducing power.
$$M{n^{2 + }} < C{l^ - } < C{r^{3 + }} < Cr$$

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. $$Zn\left( s \right)$$
B. $$Cr\left( s \right)$$
C. $${H_2}\left( g \right)$$
D. $$F{e^{2 + }}\left( {aq} \right)$$
View Answer
Releted Question 2

Faraday’s laws of electrolysis are related to the

A. atomic number of the reactants.
B. atomic number of the anion.
C. equivalent weight of the electrolyte.
D. speed of the cation.
View Answer
Releted Question 3

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 :

A. $$Ag,Hg,Cu,Mg$$
B. $$Mg,Cu,Hg,Ag$$
C. $$Ag,Hg,Cu$$
D. $$Cu,Hg,Ag$$
View Answer
Releted Question 4

The electric charge for electrode deposition of one gram equivalent of a substance is :

A. one ampere per second.
B. 96,500 coloumbs per second.
C. one ampere for one hour.
D. charge on one mole of electrons.
View Answer

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Electrochemistry

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