Question

The electrode potential of \( \mathrm{M^{2+}/M} \) of 3d-series elements shows positive value for :

• A. Fe
• B. Co
• C. Zn
• D. Cu

Hint:

Copper is the only 3d series metal that does not displace $H_2$ gas from dilute acids because its reduction potential is higher than that of $H^+$.

Answer 1

The Correct Option is D

The electrode potential in the context of the 3d-series elements refers to the standard electrode potential of various metal ions transitioning into their elemental forms. This value gives insight into the tendency of a metal ion to gain electrons and transform into the metal. It is also known as the standard reduction potential.

In the following options, we are concerned with the M^{2+}/M electrode potential for the 3d-series elements. Let's evaluate each option:

1. Fe (Iron): The standard electrode potential for \mathrm{Fe^{2+}/Fe} is approximately -0.44 \, \mathrm{V}. This is a negative value, indicating that the reduction is not favorable compared to the standard hydrogen electrode.
2. Co (Cobalt): The standard electrode potential for \mathrm{Co^{2+}/Co} is roughly -0.28 \, \mathrm{V}. This is again a negative value.
3. Zn (Zinc): The standard electrode potential for \mathrm{Zn^{2+}/Zn} is about -0.76 \, \mathrm{V}. Zinc, therefore, also shows a negative electrode potential.
4. Cu (Copper): The standard electrode potential for \mathrm{Cu^{2+}/Cu} is approximately +0.34 \, \mathrm{V}. This is a positive value, indicating that the reduction of copper ions to metallic copper is favorable.

From the above evaluations based on electrode potential values, only copper (Cu) has a positive electrode potential. This means among the given options, copper is more prone to reduction under standard conditions due to its positive value.

Conclusion: The electrode potential of \mathrm{M^{2+}/M} shows a positive value for the element Cu.