Turn in ONE copy of the quiz for all team members. Show all work and use signifi
ID: 588909 • Letter: T
Question
Turn in ONE copy of the quiz for all team members. Show all work and use significant digits and appropriate units. Draw a box around your answers when finished. 1) Consider the following balanced reaction at 25.0 °C: 2Mno,"(aq) + 16H. (aq) + 5pb (s) + 5so,"(aq) 5pbso, { s) + 2Mn2. (aq) + 8H20 ( 1 ) la) Determine the cell potential under the following conditions at 25 C? (5 pts) [MnO] = 0.600 M [SO4 2] = 1.200M [Mn2+] = 1.00 x 10"M pH = 1.60 lb) If, concentrated HCI was added to the system, would the cell potential increase or decrease? Explain your answer! 2 pts Ic) What is the value of the equilibrium constant at 25 °C? Does the reaction favor products or reactants? (3 pts)Explanation / Answer
Q!
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
E°cell = Ered - Eox
MnO4 + 8 H+ + 5 e Mn2+ + 4 H2O +1.51; (acthode)
PbSO4(s) + 2 e Pb(s) + SO42 0.3588 (anode)
E°cell = 1.51+0.3588 = 1.8688 V
Q = [Mn+2]^2 /([MnO4-]^2 * [H+]^16 * [ SO4-2]^5
[H+] = 10^-pH = 10 ^-1.6
Q = (10^-4)^2 / ((0.6^2)(10 ^-1.6)^6 * (1.2^5))
Q = 44.44
Ecell = E° - (RT/nF) x lnQ
Ecell = 1.8688 - (8.314*298)/(10*96500) * ln(44.44)
Ecell = 1.859 V
b)
if HCl is added, then H+ increases, which is a reactant so E cel lmust increase since this favours reaction forward
c)
d G= -n*F*Ecell
dG = -10*96500*1.859
dG =-1793935 J/mol
dG = -1793.935 kJ/mol
dG = -RT*ln(K)
-1793935= -RT*ln(K)
K = exp(1793935/(8.314*298))
K = exp(724.069)
this favours strongly towards products
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