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.