Consider the following electrochemical cell notation: M(s) / M2+ (aq)// H+ (aq)/ H2(g). The value of the electrode potential is positive when?

Analyze the question

The cell is written as M(s) | M²⁺(aq) || H⁺(aq) | H₂(g). By convention, the left-hand electrode is the anode (oxidation) and the right-hand electrode is the cathode (reduction). The question asks when the measured cell potential (E°cell) is positive.

Concept

E°cell = E°(right) − E°(left) = E°(cathode) − E°(anode)

Here the hydrogen electrode (E° = 0.00 V) is on the right, so:

E°cell = 0 − E°(M²⁺/M) = −E°(M²⁺/M)

For E°cell to be positive, E°(M²⁺/M) must be negative. That means M is more reactive than hydrogen and readily loses electrons.

Solution

1. M is oxidised at the left electrode: M(s) → M²⁺(aq) + 2e⁻
2. The electrons travel through the external wire from M to the hydrogen electrode.
3. At the hydrogen electrode, H⁺ is reduced: 2H⁺(aq) + 2e⁻ → H₂(g)

So the cell potential is positive, and the cell is spontaneous as written, when electrons flow from the metal electrode M(s) to the hydrogen electrode.

Checking the other options

• B: Electrons flowing from H₂ to M gives a negative potential as written, because the reaction would be reversed.
• C: A high current does not determine the sign of the potential.
• D: At equilibrium there is no net flow, so the cell potential is zero.

Answer: A

Tip: In cell notation, the left side is oxidation, so electrons leave from the left and go to the right. If the cell is written this way and electrons really do flow left to right, E°cell is positive. Remember "LEFT = anode, electrons leave."