Question

Two beakers containing different buffer solutions of lactic acid / lactate (CH3CHOHCO2H / CH3CHOHCO2 - , or LacH / Lac- ) and quinhydrone are prepared. In one beaker the [LacH] = 0.25 M and the [Lac- ] = 0.50 M, in the second beaker the [LacH] = 0.50 M and the [Lac- ] = 0.25 M. Given that the Ka of LacH is 4.4 x10-4 , what is the voltage produced when these two cells are connected?

Answer 1

(i) pKa of LacH = -log (4.4 X 10^-4 )

..........................= 3.3565

(ii)

a)Calculate pH of the solutions in two beakers containing qiinhydrone separately using Handerson's equation-

pH = pK_a   + log ( [ Lac ^- ] / [ LacH ] )  

b) and there from calculate the respective [H^+]

Beaker 1 ,

pH = 3.3565 + log (0.50 /.25 )

...... = 3.3565 + 0.3010

........= 3.6575

& [ H^+ ] = antilog ( - 3.6575 )

................. = 2.200 x 10^-4

similarly for second beaker ,

pH = 3.3565 + log ( 0.25 / 0.50 )

...... = 3.3565 + log (0.5 )

........= 3.3565 - 0.3010

........= 3.0555

& [ H^+ ] = antilog (-3.0555)

..................= 8.800 x 10^-4

(iii) The potential ,E _Q  of the quinhydrone electrode depends on the conentration of H^+ ions in solution. As worked out from Nernst equation . it is given by the relation

E_Q   = E^o_Q   - { [2.303( RT / F ) ] log [ H⁺ ] }

or ..= E^o_Q   + { [ 2.303 ( RT / F ) ] pH.......................................(1)

The standard reduction potential E^o_Q , of the quin hydrone electrode is known to be 0.6996 Thus from of (1) we have

E_Q   = 0.6996 + 0.0591 x pH..............................................(2)

Calculate E_Q for the two beakers separately using (2)   

Beaker 1 ....................E_Q   = 0.6996 + 0.0591 x 3.6575

.............................................= 0.9157 V

Beaker 2......................E_Q   = 0.6996 + 0.0591 x 3.0555

.............................................= 0.8802 V

(iv) So when the two beakers are connected a voltage produced = ( 0. 9157 - 0.8802 )

.......................................................................................................= 0.0355 V