A cell reaction involving 2 electrons has ((partial E deg)/(partial T))_P = 2.0 x 10^-4 V K^-1. Taking F = 96500 C mol^-1, Delta_r S deg is

NEET levelSingle correct optionDifficulty 1 of 3ec056

A cell reaction involving 2 electrons has $(\frac{\partial E ^{\circ}}{\partial T})_{P} = 2.0 \times 1 0^{- 4} V K^{- 1}$ . Taking $F = 96500 C mo l^{- 1}$ , $Δ_{r} S^{\circ}$ is

Δ_{r} S^{\circ} = n F (\frac{\partial E ^{\circ}}{\partial T})_{P} = 2 \times 96500 \times 2.0 \times 1 0^{- 4} = 38.6 J K^{- 1} mo l^{- 1}

. The sign follows the sign of the temperature coefficient, so a positive coefficient gives positive entropy change. Option B forgets

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