Consider the above electrochemical cell where a metal electrode ( M ) is undergoing redox reaction by forming M⁺(M→ M⁺+e⁻). The cation M⁺is present in two different concentrations c₁ and c₂ as shown above. Which of the following statement is correct for generating a positive cell potential?

Solution

Chemistry ma’am · 3 steps

Same metal on both sides, so only concentration drives it: the richer solution must sit at the cathode.

1. Concentration cell. Both electrodes are the same metal M in M⁺ solution, so E°_(cell) = 0. At the anode M → M⁺ + e⁻ adds ions. At the cathode M⁺ + e⁻ → M removes ions. The net change is M⁺(cathode side) → M⁺(anode side).

   E_(cell) = 0 - RT/F ln c_(anode)/c_(cathode) = RT/F ln c_(cathode)/c_(anode)

2. Sign of E. E_(cell) > 0 only when c_(cathode) > c_(anode). The cell runs spontaneously by moving ions from the concentrated side to the dilute side.

   Option	Cathode	Anode	E_(cell)
   (a) c₁ at anode, c₁ > c₂	c₂ (smaller)	c₁ (larger)	negative
   (b) c₁ at cathode, c₁ > c₂	c₁ (larger)	c₂ (smaller)	positive
   (c) c₁ at cathode, c₁ < c₂	c₁ (smaller)	c₂ (larger)	negative
   (d) c₁ at anode, c₁ = c₂	c₂	c₁, equal	zero

3. So. The cell potential is positive only when the cathode holds the higher concentration, that is, c₁ at the cathode with c₁ > c₂. The answer is (b).

The idea to keep. In a concentration cell E° = 0. The dilute solution is the anode and the concentrated one is the cathode, because ions flow from concentrated to dilute.

Answer from the NTA final answer key. Solution by PrepWiser.