. Consider the following half-reactions: Ag+(aq) + e– → Ag(s) E = +0.80 V
Cu2+(aq) + 2 e– → Cu(s) E = +0.34 V
Pb2+(aq) + 2 e– → Pb(s) E = –0.13 V
Fe2+(aq) + 2 e– → Fe(s) E = –0.44 V
Al3+(aq) + 3 e– → Al(s) E = –1.66 V
Which of the above metals or metal ions will reduce Pb2+(aq)?
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)

d. Fe(s) and Al(s)

In the redox reaction, it is also known  

Reducing agents are substances that experience oxidation  

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series  

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

The electrodes which are easier to reduce than hydrogen (H), have E cells = +

The electrodes which are easier to oxidize than hydrogen have a sign E cell = -

So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)

The metal  : d. Fe(s) and Al(s)

i believe the answer is true.