By Sunil Bhardwaj

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In the previos video we have already derived the relation between emf of cell and heat of reaction, as $$-nFE = \Delta H - nFT{ \left( \frac { dE }{ dT } \right) }_{ P }$$

(1) If $$\frac { dE }{ dT }$$ is positive i.e. $$\frac { dE }{ dT } > 0$$. This means emf of cell increase with the temperature. Hence electrical energy will be greater than heat of reaction $$(\Delta H)$$. Therefore extra energy must be supplied to cell from surroundings otherwise the temperature of cell will fall down and cell will be cold.

(2) If $$\frac { dE }{ dT }$$ is negative i.e.$$\frac { dE }{ dT } < 0$$. This means emf of cell decrease with the temperature. Hence electrical energy is less than heat of reaction $$(\Delta H)$$. If the heat is not removed from the cell, then the cell will become hot.

(3) If $$\frac { dE }{ dT } = 0$$. This means heat of reaction is completely converted into the emf of cell. Thus the cell will become neither cold nor hot.

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