The concentrations of NaCl and KI are at 0.160 M and 0.120 M, respectively, in t
ID: 539818 • Letter: T
Question
The concentrations of NaCl and KI are at 0.160 M and 0.120 M, respectively, in the following electrochemical cell:
Cu(s) | CuI(s) | I–(aq) || Cl–(aq) | AgCl(s) | Ag(s)
Please solve and show working.
The concentrations of NC and Kl are at 0.160 M and 0.120 M, respectively, in the following electrochemical cell Cu(s) Cul(s) | r(aq) Il Cr(aq) | AgCI(s) | Ag(s) A) Using the following half-reactions, calculate cell voltage: Cu(s)+1- E"=-0.1 Eo=0.222 V Number B) Now, calculate the cell voltage using the following half-reactions. The solubility products (Ksp) for AgCI and Cul are 1.8x10-10 and 1.0x102 respectively. Cu+ + e- Cu(s) E°=0.518 V Ag+ + e- Ag(s) Eo=0.7993 VExplanation / Answer
When the cell is NOT under standard conditions, i.e. 1M of each reactants at T = 25°C and P = 1 atm; then we must use Nernst Equation.
The equation relates E°cell, number of electrons transferred, charge of 1 mol of electron to Faraday and finally, the Quotient retio between products/reactants
The Nernst Equation:
Ecell = E0cell - (RT/nF) x lnQ
In which:
Ecell = non-standard value
E° or E0cell or E°cell or EMF = Standard EMF: standard cell potential
R is the gas constant (8.3145 J/mol-K)
T is the absolute temperature = 298 K
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant = 96485.337 C/mol or typically 96500 C/mol
Q is the reaction quotient, where
Q = [C]^c * [D]^d / [A]^a*[B]^b
pure solids and pure liquids are not included. Also note that if we use partial pressure (for gases)
Q = P-A^a / (P-B)^b
substitute in Nernst Equation:
Ecell = E° - (RT/nF) x lnQ
Get E°cell
E°cell = E-Cathode - E-Anode = 0.222--0.185 = 0.407 V
n = 1 e- transfers, T = 298K
Q = [Cl-]/[I-]
substitute
Ecell = E° - (RT/nF) x lnQ
Ecell = 0.407 - (8.314*298/(1*96500)) *ln(0.16/0.12)
Ecell = 0.39961 V
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