Question

Consider the following half-reactions:

Ag+ (aq) + e- --> Ag(s)

E cell = 0.80 VV

Cu2+(aq) + 2e- --> Cu(s)

E cell = 0.34 V

Pb2+(aq) + 2e- --> Pb(s)

E cell = -0.13 V

Fe2+(aq) + 2e- --> Fe(s)

E cell = -0.44 V

Al3+ (aq) + 3e- --> Al(s)

E cell = -1.66 V

Which of the above metals or metal ions will oxidize Pb(s)?

a. Ag+(aq) and Cu2_(aq)

b. Ag(s) and Cu(s)

c. Fe2+(aq) and Al3+(aq)

d. Fe(s) and Al(s)

e. Cu2+(aq) and Fe2+(aq)

Answer 1

Answer : option a) Ag+(aq) and Cu2+(aq)

Explanation :

Ag+ (aq) + e- --> Ag(s)   E cell = 0.80 VV

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

Pb2+(aq) + 2e- --> Pb(s)       E cell = -0.13 V

Fe2+(aq) + 2e- --> Fe(s)        E cell = -0.44 V

Al3+ (aq) + 3e- --> Al(s)          E cell = -1.66 V

Pb(s) will be oxidized , when the reduction potential of the metals greater than the reduction potential of Pb(s).

here Ag+ and Cu2+ have greater reduction potential than Pb. so these two are oxidize Pb(s)