The potential for the given half cell at 298 K is - ............ × 10 - 2 V . 2 H ( aq ) + + 2 e - → H 2 ( g ) H + = 1 M , P H 2 = 2 atm (Given 2 . 303 RT / F = 0 . 06 V , log 2 = 0 . 3 )

Step-by-step solution

The Nernst equation for the given cell reaction 2 H + + 2 e - → H 2 is E = E ° - 0 . 06 2 × log p H 2 H + 2 E = 0 - 0 . 06 2 log 2 1 2 E = - 0 . 03 × 0 . 3 = - 0 . 009 = - 9 × 10 - 3 E = - 0 . 9 × 10 - 2 V