Question

Assuming standard conditions, and considering the table of standard reduction potentials for half-reactions, given in your text, rank the following species according to their relative strength as reducing agents. For example, the most powerful reducing agent would be given rank "1", and the least "6".

Cd

Mn

H2 (acidic half-cell solution)

Na

Cl-

ClO2-

Answer 1

I do not have access to your text; however, I am using the values that are available from internet sources.

Element/Molecule/Ion	Cathode Half Reaction (Reduction)	Standard electrode potential (V)
Cd	Cd2+ (aq) + 2 e- ----> Cd (s)	-0.40
Mn	Mn2+ (aq) + 2 e- ----> Mn (s)	-1.18
H2 (acidic half-cell solution)	2 H+ (aq) + 2 e- -----> H2 (g)	0.00
Na	Na+ (aq) + e- ----> Na (s)	-2.71
Cl-	Cl2 (g) + 2 e- ----> 2 Cl- (aq)	1.36
ClO2-	ClO2- (aq) + H2O (l) + 2 e- ----> ClO- (aq) + 2 OH- (aq)	0.59

Look at the values of the standard electrode potentials. By default, standard electrode potentials are reduction potentials. The more negative the value of the reduction potential (or the lower the value of the reduction potential), the more reducing is the element or the ion. Alternatively, the more positive the value of the reductionj potential, the more oxidising will be the element or ion.

The only point we must note is that the standard reduction potential for Cl₂ is easily available from tables as +1.36; however, we are supplied Cl^-. The standard electrode potential for Cl^- will be the inverse of the standard electrode potential for Cl₂. The standard electrode potential is thus -1.36.

Now, arrange the given substances in increasing order of standard electrode potential (starting with the most negative – assign 1 here as this substance will be the most reducing). The table is as below:

Substance	Standard eletrode potential (V)	Rank (most reducing =1; least reducing = 6)
Na	-2.71	1
Cl-	-1.36	2
Mn	-1.18	3
Cd	-0.40	4
H2	0.00	5
ClO2-	0.59	6