# 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 ${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_P}.$

**Complete step by step answer:**

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}}.$**

**Additional Information:**

With the help of the temperature coefficient, we can predict the reaction that takes place inside the cell.

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 $\Delta H = - nF{E^o}.$

**Note:**

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.

The heat of a reaction would also be the amount of heat added or liberated during a chemical reaction, whereas Gibbs’s free energy could be determined by measuring the amount of electrical battery directly that the battery could do.

In simple words, Gibbs free energy is the energy associated with a chemical reaction that could be used to do work.