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 10⁶ , 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 )₆²⁻ / HSnO ₂⁻, OH ⁻ ^ =-0.90 ~V

Step-by-step solution

The electrochemical cell has cathode: Bi₂O₃ + 3H₂O + 6e^- 2Bi + 6OH^- with E° = -0.44 V and anode: Sn(OH)₆²⁻ - 2e^- HSnO₂^- + H₂O + 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).