At 298 K , the standard reduction potential for Cu 2 + / Cu electrode is 0 . 34 V . Given : K sp Cu ( OH ) 2 = 1 × 10 - 20 Take 2 . 303 RT F = 0 . 059 V The reduction potential at pH = 14 for the above couple is ( - ) x × 10 - 2 V . The value of x is

Step-by-step solution

For the reaction: Cu 2 + aq + 2 e - → Cu s The reduction potential is E = E ο - RT 2 F ln 1 Cu 2 + From the given data pH   =   14 and K sp = Cu OH 2 = 1 .0 × 10 - 20 , we get H + = 10 - 14 M , OH - = K W H + = 10 - 14 M 2 10 - 14 M = 1 M Cu 2 + = K sp OH - 2 = 1 .0 × 10 - 20 M 2 1 M = 1 .0 × 10 - 20 M E =0.34 V - 0 .059 V 2 log 1 1 .0 × 10 - 20 = 0 .34 V - 0 .059 × 20 V 2 E = 0 . 34 - 0 . 59   = - 0 . 25   V