stronger oxidizing agent?

Question

Al3+(aq) + 3e- ===> Al(s) E° = -1.66 V
I2(s) + 2e- ===> 2 I-(aq) E° = +0.54 V
Based on these data, under standard conditions,
(a) Al3+(aq) is a stronger oxidizing agent than I2(s) and I-(aq) is a strong reducing agent than is Al(s).

(b) I2(s) is a stronger oxidizing agent than Al3+(aq) and Al(s) is a stronger reducing agent than I-(aq).

(c) Al(s) is a stronger oxidizing agent than I-(aq) and Al3+(aq) is a stronger reducing agent than I2(s).

(d) I-(aq) is a stronger oxidizing agent than Al(s) and I2(s) is a stronger reducing agent than Al3+(aq).

Answer 1

Answer b is true.

Answer 2

the oxidizing agent is decreased, the oxidized species is the reducing agent to be oxidized, Cu and Zn prefer to lose electrons and the ions, Mg2+, Zn2+ and Pb2+ prefer to achieve the electrons. in the adventure that they earnings electrons, they were decreased AND oxidizing brokers. a million. Zn2+ + Cu, for Cu to be oxidized, Cu2+ should be shaped replacing the colour of the answer to blue and Zn ought to ppt out as a strong metallic, on the grounds that no rxn exceeded off, Cu(s) is more advantageous oxidizer than Zn2+ yet this isnt' declaring a lot. the actual incontrovertible fact that no rxn exceeded off tells me that neither metallic became oxidized or decreased. so, neither one is an oxidizing agent. 2. see #a million 3. Pb2+ + Zn led to Zn2+ and Pb(s) formation, Pb2+ is the more advantageous oxidizing agent in this equation on the grounds that Zn grew to grow to be Zn2+ by technique of dropping 2 electrons. Pb brought on out as a metallic and Zn2+ went into answer. universal, in accordance to easily those consequences, Pb2+ is the more advantageous oxidizing agent of the three reactions