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Use the link Standard Reduction Potentials to answer the following questions. For each pair of redox...

Use the link Standard Reduction Potentials to answer the following questions.
For each pair of redox couples, write a balanced redox equation, determine the cell potential and the number of electrons transferred. Use the lowest possible coefficients. Include states-of-matter in your answer.

a) H¹⁺/H₂ + Ag¹⁺/Ag
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E^o = V	n (moles of e^-) =
b) H¹⁺/H₂ + Pb²⁺/Pb
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E^o = V	n (moles of e^- )=
c) Al³⁺/Al + Cu²⁺/Cu
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E^o = V	n (moles of e^- )=
d) Fe²⁺/Fe + Zn²⁺/Zn
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E^o = V	n (moles of e^-) =

Ox + ne¹

Expert Solution

1)

H1+/H2 + Ag1+/Ag 0 // +0.80

Recall, the most positive will reduce, the other will oxidize, therefore

Ag+1 + 1e- ---> Ag (will reduce)

2H+ + 2e- --> H2 ( will not reduce, you need to invert this)

Balance equations

2Ag+1(aq) + 2e- + H2(g) ---> 2H+(aq) + 2e- + 2Ag(s) (aq)

E cell = Ered + Eox = +0.80 +(0) = +0.80

2 mol of electrons were transfered

2)

H1+/H2 + Pb2+/Pb 0 // -0.13

2H+ + 2e- --> H2 (will reduce)

Pb+2 +2e- --> Pb ( will not reduce, you need to invert this)

Balance Equation

2H+(aq) + 2e- + Pb(aq) ---> Pb+2(aq) + 2e- + H2(g)

E cell = Ered + Eox = 0 - (-0.13) = +0.13

2 mol of e- were transferred

3)

Al3+/Al + Cu2+/Cu -1.66 // +0.34

Cu+2 + 2e- ---> Cu(s) (will reduce)

Al+3 + 3e- ---> Al(s) (will not reduce, invert it!)

Balance Equation

3Cu+2(aq) +6e- + 2Al(s) ---> 3Cu(s) + 2Al+3(aq) + 6e-

E cell = Ered + Eox = 0.34 + 1.66 = +2.0

6 moles of electrons were transferred

4)

Fe2+/Fe + Zn2+/Zn -0.44 // -0.76

Zn+2 + 2e- ---> Zn(s) (will not reduce, invert it)

Fe+2 + 2e- ---> Fe(s) (will reduce)

Fe+2(aq) + 2e- Zn(s) ---> Fe(s) + 2e- + Zn+2 (aq)

E cell = Ered + Eox = -0.44 + 0.76 = +0.32

2 moles of electrons were transferred

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