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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

Solutions

Expert Solution

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

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

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

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

queen_honey_blossom answered 1 year ago

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