Question

NO3-(aq)+4H+(aq)+3e->NO(g)+2H2O(l) E=0.96V
ClO2(g)+e->ClO2-(aq) E=0.95V
Cu2+(aq)+2e->Cu(s) E=0.34V
2H+(aq)+2e->H2(g) E=0.00V
Pb2+(aq)+2e->Pb(s) E=-0.13V
Fe2+(aq)+2e->Fe(s) E=-0.45V

Use appropriate data to calculate E?cell for the reaction.
3Cu(s)+2NO3-(aq)+8H+(aq)->3Cu2+(aq)+2NO(g)+4H2O(l)

Express your answer using two decimal places.

Answer 1

Warning. The E values in Tables are reduction potential.

3Cu(s)+2NO3-(aq)+8H+(aq)->3Cu2+(aq)+2NO(g)+4H2O(l)

The half reactions involved are:

Cu2+(aq)+2e->Cu(s)                                          E=0.34V                                      

NO3-(aq)+4H+(aq)+3e->NO(g)+2H2O(l)           E=0.96V        

E cell = Ecathode - Eanode   (if E are both standard reduction potential, like here)

The anode is where the oxidation is produced (here at Cu/Cu2+).

Ecell = 0.96V -   0.34V =   0.62 V

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You can write also

The half reactions involved are:

Cu(s) -> Cu2+(aq)+2e          

reverse the sign for a reversed reaction     E= - 0.34V (oxidation potential)                                     

NO3-(aq)+4H+(aq)+3e->NO(g)+2H2O(l)           E= 0.96V (reduction potential)

Add the two reactions and the two potentials

E cell = -0.34 + 0.96 = 0.62 V

This is the most clear way, avoiding confusion.

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You can also write

E cell = Ecathode +Eanode   , but Ecath is from table and Eanode is from table but with changed sign to become oxidation potential.

Ecell = 0.96V +(- 0.34V) =   0.62 V