By Sunil Bhardwaj

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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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