28. Calculate the cell voltage for the following reaction: Cu\'. (0.010 M ) + H2

Question

28. Calculate the cell voltage for the following reaction: Cu'. (0.010 M ) + H2(l atm) Cu(s) + 2H'( pH = 7.0) A) 0.19Vv B) -0.01 V C) 0.34 v D) 0.69Vv E) 0.49 V 16 29. Which one of the following is consistent with a galvanic cell? 30. How many coulombs would be required to electroplate 35.0 grams of chromium by passing an electrical current through a solution containing CrCl,? A) 6.50 ×104 C B) 2.16 x 10 C C) 6.40 x 10* C D) 1.95 x 10 C E) 1.01 x 10 C 31. Beta particles are identical to A) protons. B) helium atoms. C) hydrogen atoms. D) helium nuclei. E) electrons. 32. List the number of protons, neutrons, and nucleons (protons + neutrons), in that order, for an isotope with the symbol: Cs A) 137, 55, 192 B) 55, 137, 192 C) 55, 82, 137 D) 82,55, 137 E) 82, 137, 219

Explanation / Answer

28. pH = - log H+
H+ = 10 ^ ( - pH )
H+ = 10 ^ ( - 7.0 )
H+ = 1x10-7 mol/L of 2H+

Nernst equation
E= Eo - [ (RT) / (nF) ] ln [ Q ]
E=cell emf at diff temp and molarity
Eo=cell emf measured at standard 1 atm, 289K, 1M
R= gas constant 8.314 472 J/mol/K
T=temp in K
n=no. of electrons transferred
F=faraday constant 96 500 C/mol
Q=products ions molar concentration / reactants ions molar concentration

reactants ---> products
Cu2+ + H2 ---> Cu + 2H+
[ a ][ A ] + [ b ][ B ] ---> [ c ][ C ] + [ d ][ D ]
capital letter is the molarity of species
small letter is the coefficient of species
[ (A^a)(B^b) ] ---> [ (C^c)(D^d) ]
the molarity is equal to one for non-ionic species
example Cu, molarity= 1

writing the half-cell equation
oxidation at anode
reduced species ---> oxidized species + electrons
H2 ---> 2H+ + 2e-
[ (A^a) ] ---> [ (C^c) }
[ (1^1 ) ] ---> [ (1x10-7 ^ 2) ]
Eo=0 from standard emf table
normally to find the anode emf from the table, you have to reverse the ionic equation and reverse the sign of emf. but for the cathode emf, just copy the Eo directly from the table
E = Eo - [ (RT) / (nF) ] ln [ Q ]
E = 0 - [ (8.314*298) / (2*96500) ] ln [ (1x10-7)^2 / (1^1) ]
E = 0.414V

reduction at cathode
oxidized species + electrons ---> reduced species
Cu2+ + 2e- ---> Cu
[ (A^a) ] ---> [ (C^c) ]
[ (0.01 ^ 1) ] ---> [ (1 ^ 1 ) ]
Eo = 0.34v from standard emf table
E = Eo - [ (R*T) / (e*F) ] ln [ Q ]
E = 0.34 - [ (8.314*298) / (2*96500) ] ln [ (1 ^ 1) / (0.01^1) ]
E = 0.281V

cell emf = anode emf + cathode emf
cell emf = 0.414 + 0.281
cell emf = 0.695V = 0.69V

Answer is D

30. Chromium carries +2 charges and reduces to Mg according to
Cr+3 + 3e- ---> Cr

One mole Cr+3 ions requires 3 moles of electrons to form 1 mole of Cr.

So, now you calculate the number of moles of electrons needed to deposit 35g of Cr
(35g Cr) x (3 moles/ 51.9961g Cr) = 2.02

Therefore, C= (2.02) x (9.647 x 104) = 1.95x105 C

Answer is D

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