On the basis of the electrochemical series, which of these ions will show the greater tendency to be discharged at the cathode in an electrolytic cell
cu\(^{2+}\)
sn\(^{2+}\)
fe\(^{2+}\)
zn\(^{2+}\)
Explanation
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Discussions (13)
Cathode is a negative electrode and therefore attracts positive ions. All the ions listed in A-D all have a tendency to be discharged at the cathode because they are positive ions but cu2+ has a greater tendency because it is lower in the electrohemical series than the other electropositive elements/ions given as options in this question
The lower you are in the series as a positive ion the higher the tendency of you being discharged.
To determine which ion will show the greater tendency to be discharged at the cathode in an electrolytic cell, we need to refer to the electrochemical series (also known as the reduction potential series).
The standard reduction potentials for the given ions are:
- Cu²⁺: +0.34 V
- Sn²⁺: -0.14 V
- Fe²⁺: -0.44 V
- Zn²⁺: -0.76 V
The more positive the reduction potential, the greater the tendency to be discharged at the cathode.
Therefore, the correct answer is:
A. Cu²⁺
Cu²⁺ has the most positive reduction potential among the given ions, indicating that it will show the greatest tendency to be discharged at the cathode in an electrolytic cell.
Myschool
whyyyyyyyyy
The f**kin answer is cu2+
The explanation you showed us talking about electrode potential.
And the higher the electrode potential, the less likely it is to be discharged.
In that question, copper has the highest tendency to be discharged bcoz ut has the lowest standard electrode potential btw the 4
Thanks
IT IS EITHER YOUR WRONG OR JAMB(UTME) IS WRONG .YOUR SUPPOSED ANSWER IS IRON BUT TIN ALSO REACTS WITH STEAM TO LIBERATE HYDROGEN GAS .SO THEY ARE TWO ANSWERS. WHICH SHOULD I PICK 
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