Question

Consider your observations of the halogens. Refer to the table of Standard Reduction Potentials (Links to an external site.)Links to an external site. (the table is arranged alphabetically). What can you infer about the relationship between oxidant strength and standard reduction potential? i.e. if a halogen is a strong oxidizing agent does it have a more positive or more negative standard reduction potential?

With that in mind, would you expect MnO4-(aq) to react with NaI in acidic or in basic solution?

(Hint: find the two potentials involving MnO4-, one in acidic medium and one in base.)

acidic

basic

Answer 1

Standard reduction potential can be defined as the tendency of a chemical species to be reduced. The higher is the reduction potential of an element, the higher is the oxidant strength. For example, the standard reduction potential of Fluorine is 2.87V, and oxidant strength of fluorine is the highest in the periodic table.

A negative reduction potential means the tendency to lose electrons and a positive reduction potential indicate a tendency to gain electrons. Since halogens are a strong oxidising agent(tendency to gain electrons/electronegative), it has more positive standard reduction potential.

Since MnO₄^- is not stable it will be converted to MnO₄^2- under both acidic and basic conditions

MnO₄^-(aquous)+ 1e^- MnO₄^2- (E₀= 0.56V)

Under acidic conditions,

MnO₄^2-+4H⁺+2e^-​MnO₂+2H₂O (E₀=2.26V)

Therefore, E₀= E(reduced state)-E (oxidised state)

E₀= 2.26-0.56 =1.70 V

Under basic conditions,

MnO₄^2-+ 2H₂O+2e^-MnO₂+4OH^- (E0= 0.67V)

There fore, E₀= 0.67-0.56V= 0.11 V

Since the potential involving MnO₄^- in acidic condition is much higher than that of MnO₄^- in basic solution, MnO₄^- under acidic condition can act as a stronger oxidant and react with NaI