Answer
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Hint: The transport number is defined as the total current carried by an ion. It is also called a Hittorf number or transference number. It can be expressed as ${{\rm{t}}_{\rm{ + }}}$or ${{\rm{t}}_ - }$ or ${{\rm{t}}_{\rm{c}}}$and ${{\rm{t}}_{\rm{a}}}$ or ${{\rm{n}}_{\rm{c}}}$.
Complete step by step answer:
Given,
The fall of concentration around the anode = 0.0005124 equivalent
Mass of the copper deposited in the coulometer = 0.03879g
$\therefore $ We calculate the equivalent of the copper deposited in coulometer, which is,
$ = \dfrac{{0.03879}}{{31.9}} = 0.001216\,{\rm{equivalent}}$
Now, we calculate the transport number of ${{\rm{t}}_{{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}$
${\rm{ = }}\dfrac{{{\rm{No}}{\rm{.}}\,{\rm{of}}\,{\rm{equivalent}}\,{\rm{of}}\,{\rm{A}}{{\rm{g}}^{\rm{ + }}}\,{\rm{ion}}\,{\rm{lost}}\,{\rm{in}}\,{\rm{anode}}}}{{{\rm{No}}{\rm{.}}\,{\rm{of}}\,{\rm{equivalent}}\,{\rm{of}}\,{\rm{C}}{{\rm{u}}^{{\rm{2 + }}}}\,{\rm{deposited}}\,{\rm{in}}\,{\rm{coulometer}}}}$
Or the total fall in the concentration of the solution is,
$ = \dfrac{{0.0005124}}{{0.001216}}$
${{\rm{t}}_{{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}{\rm{ = 0}}{\rm{.4214}}$
Therefore, the transport number of ${\rm{A}}{{\rm{g}}^{\rm{ + }}}$is 0.4214.
Also, ${{\rm{t}}_{{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}{\rm{ + }}\;\;{{\rm{t}}_{{\rm{NO}}_{\rm{3}}^ - }}{\rm{ = }}\;1$
${{\rm{t}}_{{\rm{NO}}_{\rm{3}}^ - }}{\rm{ = }}\;{\rm{1}} - {\rm{0}}{\rm{.4214}}\;{\rm{ = }}\;{\rm{0}}{\rm{.5786}}$
Hence, the transport number of the${\rm{NO}}_{\rm{3}}^ - $ to the nearest integer will be 0.5786 $ \approx $ 6
Additional information:
The transport number of ions can be determined by moving the boundary method. Hittorf’s method, emf method and also from the ionic mobility. The temperature rise brings the transport number of the anion and the transport number of the cation closer to 0.5.
Note:
The transport number is defined as the ratio of the mobility or velocity of a given ion to the summation of mobilities or velocities of anion and cation. It dependents on the mobility of the ions in the electrolytic solution and also on the temperature of the solution.
Complete step by step answer:
Given,
The fall of concentration around the anode = 0.0005124 equivalent
Mass of the copper deposited in the coulometer = 0.03879g
$\therefore $ We calculate the equivalent of the copper deposited in coulometer, which is,
$ = \dfrac{{0.03879}}{{31.9}} = 0.001216\,{\rm{equivalent}}$
Now, we calculate the transport number of ${{\rm{t}}_{{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}$
${\rm{ = }}\dfrac{{{\rm{No}}{\rm{.}}\,{\rm{of}}\,{\rm{equivalent}}\,{\rm{of}}\,{\rm{A}}{{\rm{g}}^{\rm{ + }}}\,{\rm{ion}}\,{\rm{lost}}\,{\rm{in}}\,{\rm{anode}}}}{{{\rm{No}}{\rm{.}}\,{\rm{of}}\,{\rm{equivalent}}\,{\rm{of}}\,{\rm{C}}{{\rm{u}}^{{\rm{2 + }}}}\,{\rm{deposited}}\,{\rm{in}}\,{\rm{coulometer}}}}$
Or the total fall in the concentration of the solution is,
$ = \dfrac{{0.0005124}}{{0.001216}}$
${{\rm{t}}_{{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}{\rm{ = 0}}{\rm{.4214}}$
Therefore, the transport number of ${\rm{A}}{{\rm{g}}^{\rm{ + }}}$is 0.4214.
Also, ${{\rm{t}}_{{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}{\rm{ + }}\;\;{{\rm{t}}_{{\rm{NO}}_{\rm{3}}^ - }}{\rm{ = }}\;1$
${{\rm{t}}_{{\rm{NO}}_{\rm{3}}^ - }}{\rm{ = }}\;{\rm{1}} - {\rm{0}}{\rm{.4214}}\;{\rm{ = }}\;{\rm{0}}{\rm{.5786}}$
Hence, the transport number of the${\rm{NO}}_{\rm{3}}^ - $ to the nearest integer will be 0.5786 $ \approx $ 6
Additional information:
The transport number of ions can be determined by moving the boundary method. Hittorf’s method, emf method and also from the ionic mobility. The temperature rise brings the transport number of the anion and the transport number of the cation closer to 0.5.
Note:
The transport number is defined as the ratio of the mobility or velocity of a given ion to the summation of mobilities or velocities of anion and cation. It dependents on the mobility of the ions in the electrolytic solution and also on the temperature of the solution.
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