Consider the standard cell at 25 degrees Celsius based on the following half rea
ID: 757447 • Letter: C
Question
Consider the standard cell at 25 degrees Celsius based on the following half reactions. Pb^(2+) + 2e- > Pb Eofcell = -0.13V Zn^(2+) + 2e- > Zn Eofcell = -0.76V To the STANDARD CELL, OH-, is added to the zinc compartment causing precipitation of Zn(OH)2. After precipitation is complete, the concentration of the OH- is -0.10M and the measured cell potential is 1.05V. Calculate the Ksp value for Zn(OH)2(s). Zn(OH)2 > Zn^(2+)(aq) + 2OH-(aq) Ksp = ? PLEASE show ALL work, I rate ALL 5 stars.Explanation / Answer
First write the separate oxidation and reduction half reactions and look up the standard potential for each. Oxidation: 2Fe?² ---> 2Fe?³ + 2e? for which E° = -.77V (just because there is a 2 in front does not mean I multiply the potential by 2) Reduction: Cl2(g) + 2e? ---> 2Cl?(aq) for which E° = +1.36V Adding the two results in the standard potential for the cell being 1.36 - .77 = +.59V Now, you have to use the Nernst equation. Ecell = E° - (.0592 logQ )/ n where Q is reaction quotient and n is the number of electrons that are exchanged) Treat Q as if you are finding the equilibrium constant. It is the concentration of the aqueous /gaseous products over the aqueous /gaseous reactants with each reactant or product raised to the power indicated by their respective coefficients. If there is a gas, use the pressure instead of the concentration. Q = [Fe?³]² * [Cl?]² / ( [Cl2] * [Fe?²]² ) Plug in the values given. Q = [1.3x10?³]² * [2.2x10?³]² / ( [.56] * [1.7]² ) = 5.1 x 10?¹² In the Nernst equation: E° = +1.36V Q = 5.1 x 10?¹² n = 2 (since two electrons are exchanged; look at the half reactions) Ecell = 1.36V - (.0592 log(5.1 x 10?¹²) )/ 2 = 1.36V - (.0592 * 11.29)/ 2 = 1.36V - (.668)/ 2 = 1.36 - .334 = 1.026 = 1.03 Ecell = 1.03V