Question

Arrange Ag, Cr and Hg metal in the increasing order of reducing power. Given:
\[{{E}^{o}}_{{A{{g}^{+}}}/{Ag}\;}=+0.80V\]
\[{{E}^{o}}_{{C{{r}^{3}}}/{Cr}\;}=-0.74V\]
\[{{E}^{o}}_{{H{{g}^{3+}}}/{Hg}\;}=+0.79V\]

Answer 1

Hint: From the electrochemical series we can see that as the reduction potential increases, tendency to get reduced increases and hence oxidizing power will increase. Similarly, as the reduction potential declines, the tendency to get oxidized will increase and therefore reducing power increases. From this we can arrange the given metals in the increasing order of reducing power.

Complete step by step answer:
-We can compare the oxidative and reductive strengths of various metals or chemical species by measuring the standard potential value.
- The electrochemical series allows us to rank substances on the basis of their reducing and oxidizing power. As we know oxidation is the loss of electrons and reduction is the gain of electrons.
- The electrochemical series is constructed by positioning various redox equilibrium in order of their redox potentials or standard electrode potentials .The greatest negative ${{E}^{o}}$ values are positioned at the uppermost part of the electrochemical series, and the most positive values are positioned at the bottom.
- Keep in mind that each ${{E}^{o}}$ value will show whether the position of the equilibrium lies to the right or left of the hydrogen equilibrium. That variance in the positions of equilibrium is caused by the number of electrons upon platinum of the hydrogen electrode and the metal electrode to be different. This creates a potential difference which can be measured in terms of voltage.
- The higher the negative ${{E}^{o}}$ value, the more the position of equilibrium will lie to the left, that is the more readily the metal will lose its electrons. Hence, the more negative the value of redox potential, the stronger reducing agent the metal and the metal will have high reducing power.
- Among the given compounds Chromium has the most negative ${{E}^{o}}$ value, followed by mercury and the silver has the highest positive value.
Therefore the increasing order of reducing power is Ag < Hg < Cr.
So, the correct answer is “Option C”.

Note: It should be noted that an oxidizing agent takes electrons from the substance and reducing agent gives electrons to it. The more positive the value of ${{E}^{o}}$, the stronger will be the oxidizing power of the metal ion. It’s because the less readily the metal loses its electrons, and the more readily its ions pick them up again. Also, an oxidizing agent takes electrons from the substance and reducing agent gives electrons to it