1) Silicon oxide has a typical surface potential in an aqueous medium of ϕ,0

Transcription

1 1) Slcon oxde has a typcal surface potental n an aqueous medum of ϕ, = 7 mv n 5 mm l at ph 9. Whch concentraton of catons do you roughly expect close to the surface? What s the average dstance between two adjacent catons? What s the local ph at the surface? Soluton The bulk concentraton c of the + and l - ons wll be: = = 5 mm + cl The H + concentraton n the bulk at PH=9 s gven by: PH a 9 = log M + = + = H 9 1 H The Boltzmann for ons n the electrolyte at potental ϕ, wll determne the concentraton at x= n ( x) n z Fϕ ( x) z Fϕ ( x) ( x) = = exp = N N RT exp RT t x = ϕ ( x) = ϕ = 7 mv, () = exp z Fϕ RT, For +, : + () = + exp = mm 3.27 RT For H +, 11 : + () = + exp = mm H H RT too small

2 () + () catons ϕ ( x), ϕ ( x) The average dstance <d> can be found by calculatng the volume that each of the ons occupes at a concentraton c + () 1 unt = () V N + < d >= a = V unt 1 3 < d > + 1/ 3 < d >= N + () = 7.97 nm a 11 ompare wth r + = m =.9nm the average dstance <d> shows that the dstance between + s much larger than + ons. Ths can be a test for your calculatons. The local PH at x= ( ) ( 11 PH a + ) = log () = log = H

3 2) Supportng electrolyte: dsorpton of ons from the electrolyte soluton s mportant n studyng adsorpton phenomena on metal electrodes. Suppose we want to study specfc adsorpton of an anonc speces -. The concentraton of the salt B n the bulk of the soluton s.1 M. In addton to that, there s a supportng electrolyte D of concentraton 5M. The surface charge of the metal electrode s σ M =.3 /m 2. We wll use the Gouy-hapman theory to study the role of the supportng electrolyte. Unfortunately, ths theory can only be used drectly for a stuaton wth only one (z-z) electrolyte. Therefore, we have to proceed n several steps. a) onsder the stuaton when the supportng electrolyte s absent. alculate the surface potental ϕ,. alculate the concentraton of - at the surface. What s the rato between surface concentraton and bulk concentraton? b) Now consder the stuaton when only the supportng electrolyte s present. Perform the same calculaton as n a). What s the surface potental ϕ, n ths case? c) What s the effect of the potental dstrbuton calculated for the stuaton n b) on the dstrbuton of -? Dscuss how the supportng electrolyte works. Soluton In absent of supportng electrolyte:, = 8RT snh (*) 2RT M σ εε For our electrolyte: -,B + n water wth and σ =.3 2 : m = =, ε = 78.1M.1 mol Lt

4 2 965 J 12 mol 3 Lt.3 8(8.31 )(3 )( )(78)(.1 )(1 ) snh mol = K ϕ, m K. mol Nm Lt m J 2(8.31 )(3 K ) K. mol ϕ, =.23 V The correspondng surface concentraton of s gven by the Boltzmann law:, () = exp M (**) = RT The rato () = = RT, exp Postve sgn n the exponent means that we consder negatve charges ( - ). (b) When only supportng electrolyte s presented we get: J 12 mol 3 Lt.3 8(8.31 )(3 )( )(78)(5. )(1 ) snh mol = K ϕ, m K. mol Nm Lt m J 2(8.31 )(3 K ) K. mol V M RT sup sup,, =.51 ; () = exp = ϕ () = 7.83 (c) ddng supportng electrolyte (.e. ons n hgh concentratons whch

5 do NOT undergo reactons wth the electrode metal or electrolyte, hence reman nert) wll result n a much smaller ϕ, and that s, the concentraton of the reactant electrolyte. n more unform throughout the ddng of supportng D at 5M wll lead to a surface concentraton of of sup ().1 exp RT, = M 965.1M exp mol =.5V.71 M J = (8.31 )(3 K ) K. mol () = 7.84 nstead of 2614!! () σ metal concentraton of - n absent of supported electrolyte concentraton of - n present of supported electrolyte sup ()

6 3) From a Tafel plot, we can estmate the exchange current densty j as well as the transfer coeffcent α. t T=25, the followng current denstes were measured for a Pt H 2 H + electrode n dluted H 2 SO 4, as a functon of the appled overpotental η. η [mv] j [m/cm2] a) alculate the exchange current densty and the transfer coeffcent. b) How many electrons are transferred per second and per cm 2 of catalyst through the metal soluton nterface at equlbrum potental. Soluton (a) Usng Tafel approxmaton αf log( j) = log( j ) + η 2.33RT We can calculate: α from the slop of the log(current)-voltage plot exchange current densty from y-axs ntercept. onclusvely, the exchange current densty and transfer coeffcent are found to be

7 ( ) log j =.9m j = 1 = 7.94 m / cm.9 2 αf 2.33RT 2.33RT = 1.59 α = 1.59 =.62 F (b) t equlbrum potental, both the oxdaton and reducton reacton at a equal rate of j. Thus the total charge transfer across the nterface wll be 2j The total number of electrons crossng the nterface per second, n e s: n e 2 j 16 = = electron / s e