If two metallic ions of the same concentration migrate to a graphite cathode, the one that would be preferentially discharged is that with the

Analyze the question

Two metal ions of equal concentration are moving to the cathode (the negative electrode) in electrolysis. We need to know which one is discharged first. This is the idea of the preferential (selective) discharge of ions.

Concept

• At the cathode, positive ions are reduced, meaning they gain electrons: Mⁿ⁺ + ne⁻ → M
• When concentrations are equal, the ion discharged first is the one that gains electrons most easily.
• This is the ion lower in the reactivity series (more positive, or less negative, standard electrode potential).
• For example, Cu²⁺ is discharged before H⁺, and H⁺ before Na⁺. Cu²⁺ has a stronger tendency to accept electrons than Na⁺.

Evaluate the options

• A. Greater stability in solution: Stability in solution does not decide discharge. Wrong.
• B. Lower electrical charge: Charge alone does not decide it. Ag⁺ and Cu²⁺ show that charge is not the deciding factor. Wrong.
• C. Smaller mass: Mass has no bearing on discharge order. Wrong.
• D. Stronger tendency to accept electrons: This matches the definition of reduction at the cathode and a higher electrode potential. Correct.
• E. Stronger tendency to lose electrons: This describes a metal that is easily oxidised, such as K or Na. Its ions are therefore hard to reduce. Wrong.

Answer: D, the ion with the stronger tendency to accept electrons.

Exam tip: Remember "cathode = reduction = gaining electrons." The ion lowest in the reactivity series is discharged first. Order of cathodic discharge: Ag⁺ > Cu²⁺ > H⁺ > Pb²⁺ > Zn²⁺ > Na⁺ > K⁺ (concentration effects aside).