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The relation between emf of cell and heat of reaction \(\Delta H\) can be calculated with the help of Helmholtz equation. $$ \Delta G = \Delta H + T{ \left( \frac { d\left( \Delta G \right) }{ dT } \right) }_{ P }$$ Where \(\Delta G =\) free energy change, \( \Delta H =\) heat of reaction and T = temperature (K).
But $$\Delta G = -nFE\qquad ....(1) $$ $$ \therefore -nFE = \Delta H + T{ \left( \frac { d\left( -nFE \right) }{ dT } \right) }_{ P } $$ $$ -nFE = \Delta H - nFT{ \left( \frac { dE }{ dT } \right) }_{ P } $$ where \(\frac { dE }{ dT } = \frac { E_{ 2 } - E_{ 1 } }{ { T }_{ 2 } - { T }_{ 1 } }\) is rate of change of emf of cell with respect to temperature. It is also known as Temperature Coefficient. $$ \therefore -nFE = \Delta H - nFT{ \left( \frac { dE }{ dT } \right) }_{ P } $$ $$ \therefore \Delta H = nFT{ \left( \frac { dE }{ dT } \right) }_{ P } - nFE $$ $$ \therefore \Delta H = nF\left[ T{ \left( \frac { dE }{ dT } \right) }_{ P } - E \right] \qquad ....(2)$$ We know that \(\Delta G = \Delta H - T\Delta S\)
lets put the values of \(\Delta G\) and \(\Delta H\) from equation 1 and 2 $$ -nFE = nF\left[ T{ \left( \frac { dE }{ dT } \right) }_{ P } - E \right] - T\Delta S $$ $$ -nFE = nFT{ \left( \frac { dE }{ dT } \right) }_{ P } - nFE - T\Delta S $$ Or $$ T\Delta S = nFT{ \left( \frac { dE }{ dT } \right) }_{ P } $$ $$ \therefore \Delta S = nF{ \left( \frac { dE }{ dT } \right) }_{ P }\qquad ....(3)$$ With the help of equation 2 and 3 we can find out the \(\Delta H\) and \(\Delta S\) if we know \(\frac { dE }{ dT } \) i.e. temperature coefficient.
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