Consider the following electrochemical cell at 298 K Pt | HSnO _ 2 ^ - ( aq ) | Sn ( OH )_ 6 ^ 2- ( aq ) | OH ^ - ( aq ) | Bi _ 2 O _ 3 ( ~s ) Bi ( s ). If the reaction quotient at a given time is 10^ 6 , then the cell EMF ( E _ cell ) is \_\_\_\_ 10^ -1 ~V (Nearest integer). Given the standard half-cell reduction potential as E _ Bi _ 2 O _ 3 / Bi , OH ^ - ^ =-0.44 ~V and E _ Sn ( OH )_ 6 ^ 2- / HSnO _ 2 ^ - , OH ^ - ^ =-0.90 ~V
A. A
The electrochemical cell has cathode: Bi_2O_3 + 3H_2O + 6e^- 2Bi + 6OH^- with E° = -0.44 V and anode: Sn(OH)_6^ 2- - 2e^- HSnO_2^- + H_2O + OH^- with E° = -0.90 V. The standard cell EMF is E°_ cell = -0.44 - (-0.90) = 0.46 V. Using the Nernst equation with Q = 10^6 and n = 6 electrons transferred: E_ cell = 0.46 - 0.059 6 (10^6) = 0.46 - 0.059 6 6 = 0.46 - 0.059 = 0.401 V. Expressing as x 10^ -1 V: 0.401 = 4.01 10^ -1 V, so x = 4 (nearest integer).
Related: Chemistry — Electrochemistry · All PYQ Banks