Question

The best reducing agent of the following is:
a) \[\mathop H\nolimits_2 S\]
b) \[\mathop {Cl}\nolimits^ - \]
c) \[\mathop {S\mathop O\nolimits_4 }\nolimits^{2 - } \]
d) \[\mathop O\nolimits_3 \]

Answer 1

Hint: Try to recall the definition of reducing agent which says that those reactants in a chemical reaction which themselves oxidizes or loses electrons in a redox reaction.

Complete step by step solution:
As we already know that reducing agents are those spices which reduce others and themselves get oxidized by losing electrons and therefore, increasing their oxidation state in a redox reaction. Now, coming to the question in which we have to pick out the best reducing agent from the given options. It can be easily done by just keeping the definition of reducing agent in mind and inspecting each option. So, here we go-
In option A which is $\mathop H\nolimits_2 S$ , here we can see that the oxidation state of sulphur is -2 which is its minimum oxidation state and cannot be decreased below this. So, it can only lose electrons and will act as a reducing agent in a redox reaction.
In option B which is $\mathop {Cl}\nolimits^ - $, we can easily say that the oxidation state of chlorine is -1 which is also its minimum oxidation state. So, it will also act as a reducing agent.
In option C which is $\mathop {S\mathop O\nolimits_4 }\nolimits^{2 - } $ , we can easily find out that oxidation state of sulphur is +6 which is its maximum oxidation state and cannot be increased further. So, it will tend to gain electrons and will act as an oxidizing agent.
Lastly, option D which is $\mathop O\nolimits_3 $ i.e. ozone already knows that it is an oxidizing agent.
Now, between $\mathop H\nolimits_2 S$ and $\mathop {Cl}\nolimits^ - , \mathop H\nolimits_2 S$ will be stronger reducing agent as sulphur has stronger tendency to lose electron due to less ionization energy compared to chlorine.

Therefore, from above we can now conclude that option A is the correct answer.

Note: In order to find out the reducing agent, we have to compare by calculating the oxidation state and also, many times we can compare by knowing their standard reduction potential.