18) A concentration cell is assembled with a copper alol 0.25 M and the other ha

Question

18) A concentration cell is assembled with a copper alol 0.25 M and the other has [Cu2+] = 2.5 M. What is the cell potential? D) 0.0592 V A) 0 V B) 0.030 V C)-0.030 E)-0.0592 v 19 Determine the cell notation for the redox reaction given below. Fe3+(aq) 3 Cl2(g) + 2 Fe(s) 6 Cl-(aq) + 2 A) CI28)I Cr(ag)I Pt ll Fets) I Fe3+(aq) B) Cl(aq) I Cl2g)I PtlI Fe3+(ag) I Fe(s) C) Fe3+(aq) I Fe(s)II Cl (aq)I CI2(g) 1 Pt D) Fe(s) 1 CI2(g) II Fe3+(aq) I C(aq)I P E) Fefs) I Fe3+(aq) Il Cl2(g) I Cl(aq) I Pt 20) Use the tabulated half-cell potentials to calculate Go for th reaction. Cu2+(aq) + Ph(s) Cu(s) + Pb2+(aq) (Pb/Pb)-0.13 V,E (Cu/Cu)-+0.34 V A)-41 k) D)-91 k B)-0A7 K) C) +46 vJ E) +91 k oras question How many seconds are vequired to produce 4.00 g of alumi moten Aicty with an electrical current of 12.0 A7 5 23

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

E°cell = Ecu - Ecu = 0 , sinc esame Cupper

Ecell = 0 - (8.314*298/(2*96500) *ln([Cu+2]ox/[Cu+2]red)

Ecell = 0 - (8.314*298/(2*96500) *ln(0.25/2.5)

Ecell = 0.02955

Ecell = 0.30 V

Q19

oxidation // reduction

then, ignore Cl2 in a,b,

also, note that Fe+3 starting is not valid

possible naswers are D and E

ignore D since Fe(s)/ Cl2(g) can't be in the same oxidation half cell

therefore

this must be E

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