Consider the following cell reaction Cd ( s ) + Hg 2 SO 4 ( s ) + 9 5 H 2 O ( l ) ⇌ CdSO 4 · 9 5 H 2 O ( s ) + 2 Hg ( l ) . The value of E cell 0 is 4 . 315 V at 25 ° C . If ΔH ° = - 825 . 2 kJ mol - 1 , the standard entropy change ΔS ° in J K - 1 is _________ . (Nearest integer) [Given : Faraday constant = 96487 C mol - 1 ]

Step-by-step solution

For given cell reaction n = 2 and ΔG ° = - nFE ° , cell   ΔG ° = ΔH ° - TΔS Then ΔH ° = 825 . 2 × 10 3   J / mole T = 298   K E ° cell = 4 . 315   V F = 96487   C ΔS ° = - - nFE ° cell - ΔH T ΔS ° = - - 2 × 96487 × 4 . 315 - - 825 . 2 × 10 3 298 = 832 . 682 × 10 3 - 825 . 2 × 10 3 298 = 7482 298 × 10 3 ΔS ° = 25 . 1   J / k