Cu s + Sn 2 + 0 . 001 M → Cu 2 + 0 . 01 M + Sn s The Gibbs free energy change for the above reaction at 298 K is x × 10 - 1 kJ mol - 1 . The value of x is____[nearest integer] [Given : E Cu 2 + / Cu ⊖ = 0 . 34 V ; E Sn 2 + / Sn ⊖ = - 0 . 14 V ; F = 96500 Cmol - 1 ]

Step-by-step solution

E cell ° = E Sn 2 + Sn ° - E Cu 2 + Cu ° = - 0 . 14 - 0 . 34 = - 0 . 48 E cell = E cell ° - 0 . 059 2 log Cu 2 + Sn 2 + = - 0 . 48 - 0 . 0295 log 10 - 2 10 - 3 = - 0 . 48 - 0 . 0295 = - 0 . 5095 ΔG = - nFE cell = - 2 × 96500 - 0 . 5095 = 98333 . 5 = 98 . 33 kJ = 983 . 3 × 10 - 1 kJ x = 983 . 3 Nearest   integer = 983