Use the link Standard Reduction Potentials to answer the following questions. Fo

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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.

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a) H1+/H2 + Ag1+/Ag

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Eo =  V n (moles of e-) = b) H1+/H2 + Pb2+/Pb

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Eo =  V n (moles of e- )= c) Al3+/Al + Cu2+/Cu

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Eo =  V n (moles of e- )= d) Fe2+/Fe + Zn2+/Zn

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Eo =  V n (moles of e-) =

Explanation / Answer

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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