The direct determination of extremely low concentrations of ions can be quite di
ID: 505836 • Letter: T
Question
The direct determination of extremely low concentrations of ions can be quite difficult A far simpler and common procedure based on a relatively simple physical technique uses cell potentials, which are easy to measure, and evaluates ion concentrations with help of the Nernst equation. Although the Nernst equation relates cell potentials to ion activities rather than to ion concentrations, the difference between them is usually fairly small and becomes negligible in solutions where the total ionic concentration is less than about 10^-3 M. A general silver-based voltaic cell is constructed as follows: Ag(s)| Ag^+ -Solution A|| Ag^+-Solution B|Ag(s) E_cell 1) In a first cell set-up, solution A is prepared by dissolving 125mg of silver nitrate (MW: 169.9) in 125 ml_ of distilled water, while solution B is prepared by dissolving 133mg of silver acetate (MW: 166.9) in 135 mL of distilled water. Calculate the expected cell potential E_cell. 2) In a second cell set-up, solution A is prepared by adding 125mg of silver nitrate (MW: 169.9) to 250mL of 2.50M hydrochloric acid, while solution B comprises 100 mL of AgCI04 (0.100M). The cell potential is measured as E_cell = 0.541 V. Calculate [Ag^+(aq)] of solution A. 3) Use data of the second cell set-up, and calculate a value for the solubility product of silver chloride.Explanation / Answer
1) We know that concentration of a solute (c) = number of moles of the solute (n) / Volume of the solution (V)
and number of moles of the solute (n) = mass of the solute (m) / Molar mass of the solute (M)
Using these two equations we get
concentration of a solute = mass of the solute / (Molar mass of the solute X Volume of the solution)
ie c = m / MV
For solution in beaker A
c = 0.125 g/ (0.125 L X 169.9 g mol -1)
c = 1 / 169.9 mol L-1
c = 0.0058 mol L-1
For solution in beaker B
c = 0.133 g/ (0.135 L X 166.9 g mol -1)
c = 0.0059 mol L-1
Now consider the half reactions at beaker A and beaker B
Ag (0.0058 mol L-1) ------> Ag+ + e- Oxidation
Ag+ + e- ------> Ag (0.0059 mol L-1) Reduction
Ag (0.0058 mol L-1) ------> Ag (0.0059 mol L-1) E0 = 0.0 V for Overall reaction
Nernst equation is given as,
E = E0 – (RT/ zF) X log10 (aox/ ared)
At 25 deg C, RT/F can be treated as a constant with the value of 0.0592
E = E0 – (0.0592/ z) X log10 (aox/ ared)
E = 0– (0.0592/ 1) X log10 (0.0058/ 0.0059)
E = (– 0.0592) X log10 0.983 = (– 0.0592) (-0.0074)
E = 0.0004 V
The expected cell potential is 0.0004 V