# The EMF of the cell: $Pt.{H_2}\left( {1\,atm} \right)\left| {{H^ + }\left( {aq} \right)} \right.\left\| {AgCl\left| {Cl} \right.} \right.$ is $0.27\,V$ and $0.26\,V$ at ${25^o}C$ and ${35^o}C.$ The positive value of heat of reaction occurring inside the cell at ${25^o}C(in kJ{K^{ - 1}})$ is:(Write your answer to the nearest integer).

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Hint: We can calculate the heat of the reaction using electrons transferred, Faraday’s constant, standard cell potential, temperature, and temperature coefficient.
Formula used:
We can calculate the heat of the reaction with this formula,
$\Delta H = - nF{E^o} + nFT{\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P}$
Here, the heat of the reaction is $\Delta H.$
The Faraday’s constant is $F\,\left( {96500\,C \cdot mo{l^{ - 1}}} \right).$
The standard cell potential is ${E^o}.$
The number of moles of electrons transferred is ${\text{n}}{\text{.}}$
The temperature is represented as $T.$
The temperature coefficient is ${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P}.$

The given cell reaction is
$Pt.{H_2}\left( {1\,atm} \right)\left| {{H^ + }\left( {aq} \right)} \right.\left\| {AgCl\left| {Cl} \right.} \right.$
The number of electrons transferred is one.
The value of ${E_2}$ is $0.26\,V.$
The value of ${E_1}$ is $0.27\,V.$
The temperature ${T_1}$ is ${25^o}C\left( {298\,K} \right).$
The temperature ${T_2}$ is ${35^o}C\left( {308.15\,K} \right).$
The temperature coefficient is calculated as,
${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P} = \left( {\dfrac{{{E_2} - {E_1}}}{{{T_2} - {T_1}}}} \right) \\ {\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P} = \left( {\dfrac{{0.26 - 0.27}}{{308 - 298}}} \right) = \dfrac{{ - 0.01}}{{10}}\,V\,{K^{ - 1}}\left[ {{n_{cell}} = 1} \right] \\$
Therefore, we found the value of ${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P}$
Now, we can calculate the heat of the reaction by using the equation,
$\Delta H = - nF{E^o} + nFT{\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P}$
We know the value of${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P}$, F and T
Therefore, we can substitute the value in this step
$\Delta H = - nF\left[ {{E^o}_{cell}\left( {at\,{{25}^o}C} \right) - T{{\left( {\dfrac{{\partial E}}{{\partial T}}} \right)}_P}} \right] \\ \Delta H = - \left( {1\,mol} \right)\left( {96500\,C \times mo{l^{ - 1}}} \right)\left[ {0.27\,V - 298\,K\left( { - \dfrac{{0.01}}{{10}}} \right)V\,{K^{ - 1}}} \right] \\ \Delta H = - 1\,mol \times 96500\,C \times mo{l^{ - 1}} \times 0.568\,V{K^{ - 1}} \\ \Delta H = - 54812\,J{K^{ - 1}} \\ \Delta H = - 54.8\,kJ{K^{ - 1}} \\$
Therefore, the heat of the reaction is $- 54.8\,kJ\,{K^{ - 1}}.$

If $\left( {\dfrac{{\partial E}}{{\partial T}}} \right) > 0,$ then $nF{E^o} > \Delta H,$ the process inside the cell is endothermic reaction.
If $\left( {\dfrac{{\partial E}}{{\partial T}}} \right) < 0,$ then $nF{E^o} < \Delta H,$ the process inside the cell is exothermic reaction.
If $\left( {\dfrac{{\partial E}}{{\partial T}}} \right) = 0,$ then $\Delta H = - nF{E^o}.$
There is a possibility that one can use the formula $\Delta G = - nF{E^o}_{cell},$ which is completely wrong to calculate the heat of the reaction.
The heat of reaction $\left( {\Delta H} \right)$ is the quantity of heat that is released when the reactants are mixed in a beaker and permitted to react freely.