A one-electron cell has E deg = 1.000 V at 298 K and ((partial E deg)/(partial T))_P = +1.3 x 10^-4 V K^-1. The standard enthalpy change of the cell reaction at 298 K, in kJ (two decimals), is (F = 96500 C mol^-1).

JEE Main levelNumerical answerDifficulty 2 of 3ec057

A one-electron cell has $E^{\circ} = 1.000 V$ at 298 K and $(\frac{\partial E ^{\circ}}{\partial T})_{P} = + 1.3 \times 1 0^{- 4} V K^{- 1}$ . The standard enthalpy change of the cell reaction at 298 K, in kJ (two decimals), is ( $F = 96500 C mo l^{- 1}$ ).

Numerical answer type, answer in kJ, tolerance 0.02.

Δ_{r} G^{\circ} = - n F E^{\circ} = - 96500 \times 1.000 = - 96500 J

Δ_{r} S^{\circ} = n F (\partial E^{\circ} / \partial T)_{P} = 96500 \times 1.3 \times 1 0^{- 4} = 12.545 J K^{- 1}

. Then

Δ_{r} H^{\circ} = Δ_{r} G^{\circ} + T Δ_{r} S^{\circ} = - 96500 + 298 \times 12.545 = - 96500 + 3738.4 = - 92761.6 J = - 92.76 kJ