In: Chemistry
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 | |
chemPad Help |
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Eo = V | n (moles of e-) = |
b) H1+/H2 + Pb2+/Pb | |
chemPad Help |
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Eo = V | n (moles of e- )= |
c) Al3+/Al + Cu2+/Cu | |
chemPad Help |
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Eo = V | n (moles of e- )= |
d) Fe2+/Fe + Zn2+/Zn | |
chemPad Help |
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Eo = V | n (moles of e-) = |
Ox + ne1 |
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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