Chemistry · Redox Equilibrium · NEET
The anode (where oxidation happens) is always written on the LEFT. The cathode (where reduction happens) is written on the RIGHT. NCERT follows this order strictly. A quick way to remember: the alphabet order A before C matches Anode before Cathode, and left comes before right.
A single vertical line ( | ) marks a phase boundary, that is, the place where two different phases touch, like a solid metal touching its solution. A double vertical line ( || ) stands for the salt bridge that joins the two half-cells. So Zn | Zn2+ means solid zinc dipping in Zn2+ solution, and || is the salt bridge between the two beakers.
You always read the diagram outward from the salt bridge in the centre toward the electrodes at the two ends. On the left (anode) the metal loses electrons, so we write metal then ion: Zn | Zn2+. On the right (cathode) the ion gains electrons, so we write ion then metal: Cu2+ | Cu. This keeps the two solid electrodes at the far left and far right, with both solutions next to the salt bridge in the middle.
The Daniell cell reaction is Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s). Zinc is oxidised (anode, left) and copper ion is reduced (cathode, right). So the diagram is: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Sometimes concentrations are shown, for example Zn | Zn2+(1 M) || Cu2+(1 M) | Cu.
When there is no solid metal to act as the electrode (for example a gas like H2, or ions like Fe2+/Fe3+), you add an inert conductor such as platinum (Pt). Example: Pt | H2(g) | H+(aq) on the left, or Cu | Cu2+ || Fe3+, Fe2+ | Pt on the right. The Pt is written at the outer end because it is the electrode that carries current.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Two vertical lines ( || ). One vertical line ( | ) only shows a phase boundary, such as a metal touching its own solution.
Yes, for full marks. NEET expects states: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Gases must also show pressure and ions their concentration when given.
Ecell = E(cathode, right) minus E(anode, left), that is E(right) - E(left). Writing the diagram correctly with the right species on the right side gives a positive EMF for a spontaneous cell.
Separate them with a comma, not a line, because they are in the same phase (same solution): Pt | Fe2+, Fe3+. Use Pt as the inert electrode since there is no solid metal.
Yes. If you get a positive Ecell using E(right) - E(left), the anode and cathode are on the correct sides. A negative value usually means you swapped the two electrodes.
