The voltage of the cell consisting of Li s and F 2 g electrodes is 5.92 V at standard condition at 298 K . What is the voltage if the electrolyte consists of 2 M LiF . ln ⁡ 2 = 0 .693 , R = 8 .314 JK - 1 mol - 1 and F = 96500 C mol - 1

Step-by-step solution

Now, the cell reaction is Li s + 1 2 F 2 g → Li + + F - We know that, E cell = E cell o = RT nF ln ⁡ P r o d u c t R e a c t a n t = E cell o - 2 .303 RT nF log ⁡ Li + F - = 5.92 - 2.303 × 8.314 × 298 1 × 96500 log ⁡ 2 × 2 = 5.92 - 0.059 1 × 2 log ⁡ 2 = 5.92 - 0.035 = 5.887  V