Complete the following sequence. (Enter all potentials to the nearest 0.01 V. Enter calculated values to 2 . You may have to enter some values in scientific notation.)
1. complex ion formation
potential, E\deg cell, before addition of 6 M NH3 0.43 V potential, E\deg cell, after Cu(NH3)42+ formed 0.82V
Given equation (3),
Ecell = E\deg cell -
0.05922
log
(([Cu2+ ])/([Ag+ ]2))
(3)
calculate the residual concentration of free Cu2+ ion in equilibrium with Cu(NH3)42+ in the solution in the crucible. Take [Ag+ ] to be 1 M.
[Cu2+ ] = 6.6e-14 M
2. solubility product of AgCl
(a)the Cu|Cu2+||Ag+|Ag cell from the prior exercise
potential, E\deg cell
0.43V
negative electrode
Cu
(b)
the cell with 1 M KCl present
potential, Ecell
-0.10V
negative electrode
Ag
I need the following answers! Using equation (3), calculate [Ag+ ] in the cell, where it is in equilibrium with 1 M Cl- ion.(Ecell in equation (3) will be negative if the polarity is not the same as in the standard cell.) Take [Cu2+ ] to be 1 M.
[Ag+ ] = M
Since Ag+ and Cl- in the crucible are in equilibrium with AgCl, we can find Ksp for AgCl, from the concentration of Ag+ and Cl-, which we now know. Formulate the expression for Ksp for AgCl, and determine its value.
Ksp =