Cell potential an electrochemical cell based on the following two halr- reactions. Be sure to balance the equations to determine the moles of electrons. a. Ca(s) + Cu(ag)-> Ca (ag) + Cu(s) [Ca*] = 0.625 M, [Cu*] = 0.0150M b. Na(s) + Pb**(ag)-> Na (ag) + Pb(s) [Na"] = 0.00225 M, [Pb] = 0.825 M c. Cr (ag) + Cu (ag) -> Cu(s) +Cr (ag) [Cr*] = 1.85 M, [Cr] = 0.150 M, [Cu] = 0.175 M
Cell potential an electrochemical cell based on the following two halr- reactions. Be sure to balance the equations to determine the moles of electrons. a. Ca(s) + Cu(ag)-> Ca (ag) + Cu(s) [Ca*] = 0.625 M, [Cu*] = 0.0150M b. Na(s) + Pb**(ag)-> Na (ag) + Pb(s) [Na"] = 0.00225 M, [Pb] = 0.825 M c. Cr (ag) + Cu (ag) -> Cu(s) +Cr (ag) [Cr*] = 1.85 M, [Cr] = 0.150 M, [Cu] = 0.175 M
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![5) Determine the cell potential for an electrochemical cell based on the following two half-
reactions. Be sure to balance the equations to determine the moles of electrons.
a. Ca(s) + Cu (ag)-> Ca (ag) + Cu(s)
[Ca] = 0.625 M, [Cu] = 0.0150 M
b. Na(s) + Pb*(ag) -> Na (ag) + Pb(s)
[Na] = 0.00225 M, [Pb] = 0.825 M
c. Cr*(ag) + Cu*(ag)-> Cu(s) + Cr*(ag)
[Cr*] =1.85 M, [Cr] = 0.150 M, [(Cu] = 0.175 M
d. Al(s) + Fe (ag)-> Al (ag) + Fe"(ag)
[AI] = 1.00 M, [Fe] = 0.0250 M, [Fe] = 0.0350 M
e. Mno, '(ag) + Sn2 --> Mno, (ag) + Sn*(ag)
[Mno, ') = 2.25 M, [Sn-] = 1.95 M, [Mno, 2] = 0.0150 M, [Sn") = 0.0925 M
f. H,SO,(ag) + H,O + Au-> SO. (ag) + H(ag) + Au
[H,SO,] = 0.125 M, (Au"] = 0.365 M, [SO, ') = 4.15 M, [H"]= 5.00 M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7eb09b68-f1c8-419a-aa4c-a02973a45c38%2Fa73237af-dbef-4e3f-8a86-0897166d9ac3%2Fdsnvsoh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5) Determine the cell potential for an electrochemical cell based on the following two half-
reactions. Be sure to balance the equations to determine the moles of electrons.
a. Ca(s) + Cu (ag)-> Ca (ag) + Cu(s)
[Ca] = 0.625 M, [Cu] = 0.0150 M
b. Na(s) + Pb*(ag) -> Na (ag) + Pb(s)
[Na] = 0.00225 M, [Pb] = 0.825 M
c. Cr*(ag) + Cu*(ag)-> Cu(s) + Cr*(ag)
[Cr*] =1.85 M, [Cr] = 0.150 M, [(Cu] = 0.175 M
d. Al(s) + Fe (ag)-> Al (ag) + Fe"(ag)
[AI] = 1.00 M, [Fe] = 0.0250 M, [Fe] = 0.0350 M
e. Mno, '(ag) + Sn2 --> Mno, (ag) + Sn*(ag)
[Mno, ') = 2.25 M, [Sn-] = 1.95 M, [Mno, 2] = 0.0150 M, [Sn") = 0.0925 M
f. H,SO,(ag) + H,O + Au-> SO. (ag) + H(ag) + Au
[H,SO,] = 0.125 M, (Au"] = 0.365 M, [SO, ') = 4.15 M, [H"]= 5.00 M
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