Chemistry · Electrochemistry · NEET
The formula is E°cell = E°cathode − E°anode. Both E° values are standard REDUCTION potentials, exactly as written in the table. You do NOT flip any sign yourself; the minus sign in the formula already handles the anode's oxidation. Just plug in the two reduction potentials. Example: cathode +1.33 V, anode −0.44 V gives E°cell = 1.33 − (−0.44) = +1.77 V.
By convention, the LEFT side is always the anode (where oxidation happens) and the RIGHT side is always the cathode (where reduction happens). A single line | means a phase boundary (metal touching its solution). The double line || is the salt bridge that separates the two half-cells. So in Zn | Zn²⁺ || Cu²⁺ | Cu, Zn is oxidised (anode) and Cu²⁺ is reduced (cathode).
Compare the two standard reduction potentials. The electrode with the HIGHER (more positive) reduction potential is the cathode (reduction). The one with the lower (more negative) potential is the anode (oxidation). This choice always makes E°cell come out positive, which means the cell reaction is spontaneous. This is a very common NEET trick: they give two E° values and ask you to build the cell.
No. This is the mistake most students make. Keep both potentials as reduction potentials from the table. The formula E°cell = E°cathode − E°anode already subtracts the anode value, so the sign change is automatic. If you also flip the anode's sign by hand, you subtract twice and get the wrong answer.
A positive E°cell means the reaction is spontaneous (it happens on its own in a galvanic cell) because ΔG° = −nFE°cell is negative. A negative E°cell means ΔG° is positive, so the reaction is non-spontaneous and cannot occur by itself. NEET often asks 'which reaction cannot occur' — the answer is the one that gives a negative E°cell.
'n' is the number of electrons transferred in the balanced cell reaction. Find it by balancing the two half-reactions so the electrons cancel. For Zn/Cu it is 2. For 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ it is 2. You need n for ΔG° = −nFE°cell and for the Nernst equation, but you do NOT need n to compute E°cell itself (E°cell is just cathode − anode).
The correct value of cell potential (in volt) for the reaction that occurs when the following two half cells are connected is: Fe²⁺(aq) + 2e⁻ → Fe(s), E° = −0.44 V; Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, E° = +1.33 V.
At 298 K the standard electrode potentials of Cu²⁺/Cu, Zn²⁺/Zn, Fe²⁺/Fe and Ag⁺/Ag are 0.34 V, −0.76 V, −0.44 V and 0.80 V respectively. Which of the following reactions cannot occur?
For the cell reaction 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), E°cell = 0.24 V at 298 K. The standard Gibbs energy (ΔrG°) of the cell reaction is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
They are the same thing. In standard cell notation the right electrode is the cathode and the left is the anode, so E°cell = E°right − E°left = E°cathode − E°anode. Both use reduction potentials from the table.
A single vertical line | shows a phase boundary, for example a solid metal touching its ion solution (Zn | Zn²⁺). The double vertical line || shows the salt bridge that connects and separates the two half-cells.
No. E°cell (and E° of any electrode) is an intensive property. Even if you double the whole reaction, E°cell stays the same. Only ΔG° and n change with multiplication, not the potential.
EMF is the maximum voltage a cell gives when no current is drawn (open circuit). When current flows, internal resistance lowers the reading, so the measured potential difference is a little less than the EMF.
Because that choice makes E°cell positive, which corresponds to a spontaneous reaction (ΔG° = −nFE°cell < 0). A real galvanic cell always runs in the spontaneous direction, so the stronger oxidising agent (higher E°) gets reduced at the cathode.
