A voltaic cell consists of a standard reference electrode half-cell and a Cu/Cu2
ID: 526001 • Letter: A
Question
A voltaic cell consists of a standard reference electrode half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when Ecell is 0.21 V. (See the table of standard electrode potentials.)ooo AT&T; 9:47 PM 46% webassign net Standard Electrode (Half-Cell) Potentials E (V) Half-Reaction +2.87 F20g) 2e 2F (ag) 0,(g) 2H (aa) 2e +207 Co' (ag) e +1.82 2H (agi H20,(ag) HSOs (ag) 2e PbSO4(s) 2H20 Ce (ag) Ce (ag) Mn (aa) 4H20 Au (aa) 3e Au(s) 2C1 (aa) Cr207" (aa) 14H (aa) 6e 2Cr (ag) 7H20() 133 MnOMs) 4H (ag) 2e Mn (aa) 2H2O() 1.23 O20g) 4H (aa) 4e 1.23 2Br (aa) Bry() 2e +107 No, (ag) 4 ag) 3e NO(g) 2H20) +0.96 +0.92 (ag) 2e Hg2 (ag) g2 (ag) 2e +0,85 Ag (ag) e Ag(s) +0.80 Fe' (aa) er Fe (ag) +0.77 O (R) 2H (aq) 2e H202(ag) +0.68 Mno. (ag) 2H2O) 3e Mino, (s 40H (ag) +0.59 21 (ag) +0.53 O2(g) 2H2O) 4e 40H (ag) +0.40 Cu (ag) 2e Cu(s) +0.34 AgCl(s) e Ag(s) CIT (ag) +0.22 (aa) 4H (ag) 2e Son(g) 2H2O) +0.20 Cu (ang) e Cu (ag) +0.15 Sn (aq) 2e +0.13 2H (ag) Hz(g) (ag) 2e Pb(s) -0.13 sn't (aq) Sn(s) -0.14 N20g) 5H (aa) 4e N2H5 (ag) -0.23 Ni (aa) 2e Ni(s) -0.25 Co. (ang) 2e Co(s) -0.28 PbSOu(s) HT (aa) 2e Pb(s) HSO4 (ag) Cd (ag) 2e Cd(s) Fe (aq) 2e Fe(s) -0.44 cr (aa) 3e Cr(s) -0.74 Zn (ag) 2e 0.76 2H20 2e H2(g) 20H (ag) -0,83 Mn (ag) 2e Mn(s) Al' (ag) 3e Al(s) Mg (ag) 2e Mg(s) 2.37 Na ag Na(s) 2.7 Ca (ag) 2e7 Ca(s) 2.87 sr (ag) 2e Sr(s) 2.89 Ba (aq)+ 2e Ba(s) 2.90 K (ag) e 2.93 Li (aa) e Li(s) 3.05 All values at 298 K. Written as reductions: E value refers to all components in their standard states: 1 M for dissolved species, 1 atm pressure for the gas behaving ideally the pure substance for solids and liquids.
Explanation / Answer
Solution:- Copper galf cell standard reduction potential is greater than standard reference electrode(hydrogen electrode) so the reduction of copper and oxidation of hydrogen would take place.
The two half cell reactions for anode and cathode are as follows...
Anode half cell reaction:- H2(g) -------> 2H+(aq) + 2e- E0 = 0
Cathode half cell reaction:- Cu2+(aq) + 2e- ----> Cu(s) E0 = 0.34 V
Over all cell reaction:- Cu2+(aq) + H2(g) ----> Cu(s) + 2H+(aq) E0cell = 0.34 V
From Nernst equation...
Ecell = E0cell - (0.0592/n) logQ
where Q is the reaction quotient and it is [products]/reactants] and n is the number of electrons.
Q = ([H+]2)/([Cu2+]p[H2])
where pH2 stands for pressure of H2. Since we are using standard hydrogen electrode so pressure of hydrogen gas would be 1.0 atm and concentration of [H+] is 1.0 M.
Q = (1)2/([Cu2+](1) = 1/[Cu2+]
Plugging in the values in Nernst equation..
0.21 = 0.34 - (0.0592/2) log(1/[Cu2+]
0.21 - 0.34 = -(0.0592/2) log(1/[Cu2+]
-0.13 = -0.0296 log(1/[Cu2+]
divide both sides by -0.0296
4.39 = log(1/[Cu2+]
taking antilog...
104.39 = (1/[Cu2+]
24547 = (1/[Cu2+]
[Cu2+] = 1/24547 = 4.07 x 10-5
so, the concentration of [Cu2+] is 4.07 x 10-5 M.
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