Consider the following electrochemical cell at 298 K Pt | HSnO2-(aq) | Sn(OH)62-(aq) | OH-(aq) | Bi2O3(s) | Bi(s). If the reaction quotient at a given time is 106, then the cell EMF ( Ecell) is 10-1 V (Nearest integer). Given the standard half-cell reduction potential as E Bi2O3/Bi,OH- = -0.44 V and E Sn(OH)62-/HSnO2-,OH- = -0.90 V
Answer
4
Solution
Anode (oxidation): HSnO2- + H2O + 3OH- Sn(OH)62- + 2e- ] × 3
Cathode (reduction): Bi2O3 + 3H2O + 6e- 2Bi + 6OH-
Overall: Bi2O3 + 3HSnO2- + 6H2O + 3OH- 2Bi + 3Sn(OH)62-
n = 6
E cell = E cathode - E anode = -0.44 - (-0.90) = 0.46 V
Nernst equation at 298 K:
Ecell = E cell - 0.059n Q
= 0.46 - 0.0596 106
= 0.46 - 0.059 = 0.401 V
= 4.01 10-1 V 4 10-1 V
Answer: 4