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There are number of applications of emf one of the important application is this, that we can find out the formula for silver ammonia complex. The formula for silver ammonia complex is $${ \left[ { Ag }_{ m }{ \left( { NH }_{ 3 } \right) }_{ n } \right] }^{ m+ }.$$ In this formula if we know the value of n and m then we can substitute the value and can find out the correct formula of the complex. $$ { \left[ { Ag }_{ m }{ \left( { NH }_{ 3 } \right) }_{ n } \right] }^{ m+ } \longrightarrow m{ Ag }^{ + } + n{ NH }_{ 3 } $$ On applying the law of mass action. we get, $$ K = \frac { { a }_{ { Ag }^{ + } }^{ m } \times { a }_{ { NH }_{ 3 } }^{ n } }{ { a }_{ { { \left[ { Ag }_{ m }{ \left( { NH }_{ 3 } \right) }_{ n } \right] }^{ m+ } } } } = \frac { { a }_{ { Ag }^{ + } }^{ m } \times { a }_{ { NH }_{ 3 } }^{ n } }{ { a }_{ { { Complex } } } } $$ where $$ { a }_{ { { Complex } } } = { a }_{ { { \left[ { Ag }_{ m }{ \left( { NH }_{ 3 } \right) }_{ n } \right] }^{ m+ } } } $$ As solution is dilute \(\gamma = 1, \therefore a = C\) $$ K = \frac { { a }_{ { Ag }^{ + } }^{ m } \times { a }_{ { NH }_{ 3 } }^{ n } }{ { a }_{ { { Complex } } } } = \frac { { C }_{ { Ag }^{ + } }^{ m } \times { C }_{ { NH }_{ 3 } }^{ n } }{ { C }_{ { { Complex } } } } $$ $$ \therefore { C }_{ { Ag }^{ + } }^{ m } = \frac { K \times { C }_{ { { Complex } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } $$ $$ i.e. { C }_{ { Ag }^{ + } }^{ m } = { \left( \frac { K \times { C }_{ { { Complex } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } \right) }^{ \frac { 1 }{ m } }\qquad ....(1) $$ Now if we arrange the silver plate in the above silver ammonia complex, then we can say it will be \(Ag | { Ag }^{ + }\). The oxidation reaction for this electrode is, $$ { Ag } \longrightarrow { Ag }^{ + } + { e }^{ - } $$ \(\therefore K = { a }_{ { Ag }^{ + } } = C_{ { Ag }^{ + } } \) for dilute solution.
On applying the Nernst equation, $$ \underset { Ag }{ { E }_{ Ox } } = \underset { Ag }{ { E }_{ Ox }^{ 0 } } - \frac { RT }{ F } \ln { C_{ { Ag }^{ + } } } \qquad ....(2) \because n = 1 $$ On putting the value from equation (1) to equation (2) we get, $$ \underset { Ag }{ { E }_{ Ox } } = \underset { Ag }{ { E }_{ Ox }^{ 0 } } - \frac { RT }{ F } \ln { { \left( \frac { K \times { C }_{ { { Complex } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } \right) }^{ \frac { 1 }{ m } } } $$ $$ = \underset { Ag }{ { E }_{ Ox }^{ 0 } } - \frac { RT }{ mF } \ln { { \left( \frac { K \times { C }_{ { { Complex } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } \right) } } $$ $$ = \underset { Ag }{ { E }_{ Ox }^{ 0 } } - \frac { RT }{ mF } \ln { { K } } - \frac { RT }{ mF } \ln { { \frac { { C }_{ { { Complex } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } } } \qquad ....(3) $$ Now we prepare two different concentration of the same complex. Let the solutions are saturated with \({ NH }_{ 3 }\) but the concentration of silver in the two are different, then if we apply the equation (3) to these two different concentration of complex we get, for first solution it will be, $$ { E }_{ 1 } = { E }_{ Ag }^{ 0 } - \frac { RT }{ mF } \ln { { K } } - \frac { RT }{ mF } \ln { { \frac { { C }_{ { { Complex(1) } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } } } \qquad ....(4) $$ and for second solution it will be, $$ { E }_{ 2 }={ E }_{ Ag }^{ 0 }-\frac { RT }{ mF } \ln { { K } } -\frac { RT }{ mF } \ln { \frac { { C }_{ Complex(2) } }{ { { C }_{ { NH }_{ 3 } }^{ n } } } } \qquad ....(5) $$ On taking the difference of (4) and (5) we get: $$ { E }_{ 1 } - { E }_{ 2 } = \left[ { E }_{ Ag }^{ 0 } - \frac { RT }{ mF } \ln { { K } } - \frac { RT }{ mF } \ln { { \frac { { C }_{ { { Complex(1) } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } } } \right] - \left[ { E }_{ Ag }^{ 0 } - \frac { RT }{ mF } \ln { { K } } - \frac { RT }{ mF } \ln { { \frac { { C }_{ { { Complex(2) } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } } } \right] $$ $$ { E }_{ 1 } - { E }_{ 2 } = \left[ \frac { RT }{ mF } \ln { { \frac { { C }_{ { { Complex(2) } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } } } \right] - \left[ \frac { RT }{ mF } \ln { { \frac { { C }_{ { { Complex(1) } } } }{ { C }_{ { NH }_{ 3 } }^{ n } } } } \right] $$ $$ { E }_{ 1 } - { E }_{ 2 } = \frac { RT }{ mF } \ln { { \frac { { C }_{ { { Complex(2) } } } }{ { C }_{ { { Complex(1) } } } } } } \qquad ....(6) $$ In the above equation (6) we know the concentration of two complex and we know \({ E }_{ 1 }\) and \({ E }_{ 2 }\) and we know \(\frac { 2.303RT }{ F }\) then (m) is only unknown and we can calculate.
For the determination of (n): We will again prepare the two different concentrations of complex. They have same concentration of \({ Ag }^{ + }\) but different concentrations of \({ NH }_{ 3 }\) and we will find out the potential of each electrode. Again let it be \({ E }_{ 1 }\) and \( { E }_{ 2 }\) respectively, then similarly we will get equation of $$ { E }_{ 1 } - { E }_{ 2 } = \frac { RT }{ mF } \ln { { \frac { { C }_{ { NH }_{ 3 }(1) }^{ n } }{ { C }_{ { NH }_{ 3 }(2) }^{ n } } } } = \frac { RT }{ mF } \ln { { { \left( \frac { { C }_{ { NH }_{ 3 }(1) } }{ { C }_{ { NH }_{ 3 }(2) } } \right) }^{ n } } } $$ $$ { E }_{ 1 } - { E }_{ 2 } = \frac { nRT }{ mF } \ln { { { \frac { { C }_{ { NH }_{ 3 }(1) } }{ { C }_{ { NH }_{ 3 }(2) } } } } } \qquad ....(7)$$ In the equation (7) all the terms are known except (n) and it can be determined. Suppose m = 3 and n = 4 then the formula of Silver ammonia complex will be $$ { \left[ { Ag }_{ 3 }{ \left( { NH }_{ 3 } \right) }_{ 4 } \right] }^{ 3+ }.$$
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Don’t want your columns to simply stack in some grid tiers? Use a combination of different classes for each tier as needed. See the example below for a better idea of how it all works.
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For grids that are the same from the smallest of devices to the largest, use the .col and .col-* classes. Specify a numbered class when you need a particularly sized column; otherwise, feel free to stick to
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