Question

If $E_{\dfrac{{Cl{}_2}}{{Cl - }}}^\circ = 1.36V,E_{\dfrac{{Cr{}^{3 + }}}{{Cr}}}^\circ = - 0.74V \\
  E_{\dfrac{{Cr{}_2O{}_7}}{{Cr{}^{3 + }}}}^\circ = 1.33V,E{}_{\dfrac{{MnO{}_4^ - }}{{Mn{}^{2 + }}}}^\circ = 1.51V \\
$
Among which of the following, the strongest reducing agent:
a)\[Mn{}^{2 + }\]
b)\[Cr{}^{3 + }\]
c)\[Cl{}^ - \]
d)\[Cr\]

Answer 1

Hint: Electrochemical series is a list which shows the arrangement of elements in increasing order of electrode potential values. In this series the electrode that are in contact with their ions are arranged on the basis of the standard reduction or oxidation potentials values

Complete step by step answer:
Let us first study the data given in the question;
\[Cr{}_2O_7^ - \]Is converted to \[Cr{}_2O_7^ - \]. Let us first calculate the oxidation state of the first compound that comes down to \[ + 6\]. So it means that the reduction of \[Cr{}^{ + 6}\]to \[Cr{}^{3 + }\]it takes \[1.33V\].
Same way the reduction from \[Cl{}_2\]to \[Cl{}^ - \]takes \[1.36V\].
The oxidation state of \[MnO_4^ - \]is \[ + 7\]so the reduction from \[Mn{}^{ + 7}\]to \[Mn{}^{ + 2}\]takes \[1.51V\]and the reduction from \[Cr{}^{3 + }\]to \[Cr\]takes\[ - 0.74V\].
To solve this question we will take help of electrochemical series, this series consists of elements arranged with its various reduction potential. With the help of this potential we can decide which is the strongest reducing and oxidising agent and vice and versa. In case of hydrogen the reduction potential is zero. The reaction is given as, \[2H{}^ + + 2e{}^ - \to H{}_2\]. The elements above hydrogen will be positive and below hydrogen will be negative. When we move from positive to negative, the nature of reducing agents increases. Now if we talk about this in terms of oxidising potential, the more the oxidising potential the easier the element will get oxidised hence working as a strongest reducing agent. So we have to now check which of the given has highest oxidation potential or lowest reduction potential.
The lowest value is \[ - 0.74V\].i.e. the reduction from \[Cr{}^{3 + }\]to \[Cr\]

So, the correct answer is Option B.

Note: The top of the electrochemical series consists of weaker reducing agent and the bottom consists of stronger reducing agent. In the case of oxidising agents are opposite the stronger oxidising agent at the top and weak oxidising agents at the bottom.