It is saying, \"Incorrect. You switched the concentrations in the reaction quoti

Question

It is saying, "Incorrect. You switched the concentrations in the reaction quotient in the Nernst equation. Write each half-reaction as a reduction. The reaction quotient should be products over reactants.", but I did log (.12/.0033) when calculating for the overall e cell (and added the other components)

Mapd Sapling Learning For the following electrochemical cells, calculate the potential and determine if the cell reaction is spontaneous as written at 25°C A) Cu(s)I Cu2 (0.12 M) 11 Fe2 (0.0033 M)I Fe(s) Cu /Cu Fe»,Fe=-0.440 V Number Ec.733 Not Spontaneous B) Pt(s) Sn2 (0.0065 M), Sn4* (0.15 M)II Fe (0.0038 M), Fe3 (0.13 M) I Pt(s) . = 0.154 V EFe , . ,Fr"=0.771 V Number Ecel,-11.622 Not Spontaneous There is a hint available! View the hint by clicking on the bottom divider on the divider bar again to hide the hint. Close Previous Give Up & View Solution Check Answer Next Exit Hint

Explanation / 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

then

equation:

E°Cell = Ecathode - Eanode = -0.44-0.339 = -0.779 V

now..

e- transferred = 2, F = 96500 C/mol.

Q = [Cu+2]/[Fe+2] = 0.12 / 0.0033= 36.363

Ecell = E° - (RT/nF) x lnQ

Ecell = -0.779 - 8.314*298/(2*96500) * ln(36.363)

Ecell = -0.8251 V

since negative, will not be spontaneous

B)

similarly

E°cell = Ecahtode- Eanode = 0.771 - 0.154 =  0.617 V

Sn2+ + Fe+2 = Fe+3 + Sn4+2

balance

Sn2+ + 2Fe+2 = 2Fe+3 + Sn4+

Q = [Fe+3]^2 * [Sn+4] /([Sn+2][F+2]^2)

Q = (0.13^2)(0.15) /((0.0065)(0.13^2))

Q = 23.076

now

Ecell = E° - (RT/nF) x lnQ

Ecell = 0.617 - 8.314*298/(2*96500) * ln(23.076)

Ecell = 0.5767 V

this is spontaneous as written

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