The Galvanic Cell Idea
Electrochemistry, as NCERT opens Unit 2, is the study of producing electricity from spontaneous chemical reactions and of using electrical energy to drive non-spontaneous ones. A galvanic cell — also called a voltaic cell — sits on the first side of that statement: it is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. The Gibbs energy released by the reaction is harvested as electrical work that can run a motor, a heater or a fan.
The defining trick is spatial separation. In an ordinary beaker, oxidation and reduction occur in direct contact and the energy dissipates as heat. A galvanic cell physically separates the two half-reactions into two compartments, forcing the electrons to travel through an external wire. That stream of electrons through the wire is the electric current the cell delivers.
The zinc electrode dissolves and feeds electrons to the wire; copper ions plate out on the copper electrode. The salt bridge keeps each solution electrically neutral.
The Daniell Cell, Built From Scratch
The Daniell cell is NCERT's worked example throughout Unit 2. A zinc rod dips into zinc sulphate solution; a copper rod dips into copper sulphate solution. The two beakers are joined externally by a metallic wire through a voltmeter, and internally by a salt bridge. The overall reaction is:
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$
When the concentrations of $\ce{Zn^2+}$ and $\ce{Cu^2+}$ are both unity (1 mol dm⁻³), this cell develops a potential of 1.1 V. NCERT also notes the reversibility experiment: apply an opposing external voltage and increase it slowly. While $E_{\text{ext}} < 1.1\ \text{V}$ the cell keeps working, zinc dissolving and copper depositing. At $E_{\text{ext}} = 1.1\ \text{V}$ exactly, no current flows and no reaction occurs. Push beyond 1.1 V and the reaction reverses — the device now behaves as an electrolytic cell.
| Applied voltage | Electron flow | What happens at the metals |
|---|---|---|
E_ext < 1.1 V | Zn rod → Cu rod (current Cu → Zn) | Zn dissolves at anode; Cu deposits at cathode — galvanic operation |
E_ext = 1.1 V | No electron or current flow | No net chemical reaction — equilibrium balance |
E_ext > 1.1 V | Cu → Zn (current Zn → Cu) | Zn is deposited; Cu dissolves — electrolytic reversal |
Half-Cells, Anode and Cathode
The overall Daniell reaction is the sum of two half-reactions, each confined to one compartment:
$$\ce{Cu^2+(aq) + 2e^- -> Cu(s)} \quad \text{(reduction, on Cu electrode)}$$ $$\ce{Zn(s) -> Zn^2+(aq) + 2e^-} \quad \text{(oxidation, on Zn electrode)}$$
These two compartments are the half-cells, also called redox couples. The copper electrode is the reduction half-cell; the zinc electrode is the oxidation half-cell. NCERT stresses that innumerable galvanic cells can be assembled by pairing different half-cells, each consisting of a metallic electrode dipped in an electrolyte.
The naming is fixed by what occurs there, never by sign alone. The electrode where oxidation happens is the anode; the electrode where reduction happens is the cathode. In a galvanic cell the anode carries a negative potential with respect to its solution and the cathode a positive one, so electrons flow from the negative anode to the positive cathode through the wire. Conventional current flows opposite to electron flow.
Anode polarity flips between cell types
In a galvanic cell the anode is the negative terminal and the cathode positive. In an electrolytic cell the anode is positive and the cathode negative. Students who memorise "anode = positive" lose marks the moment the question switches cell type.
Anchor on the chemistry, not the sign: oxidation → anode, reduction → cathode, always.
Electrode Potential at the Interface
Why does a potential develop at all? NCERT gives a clean two-tendency picture of the electrode–electrolyte interface. Metal ions in the solution tend to deposit onto the metal, trying to make it positive. Simultaneously, metal atoms tend to leave the electrode as ions, depositing electrons behind and trying to make it negative. At equilibrium a separation of charge is established, and a potential difference develops between the electrode and the electrolyte. This is the electrode potential.
When all the species in a half-cell are at unit concentration (and gases at one bar), the value is called the standard electrode potential, denoted $E^\circ$. By IUPAC convention these are reported as standard reduction potentials: every tabulated half-reaction is written as a reduction, $\ce{M^n+ + ne^- -> M}$, and its $E^\circ$ measures the tendency of that species to be reduced.
The balance of the two opposing tendencies fixes the equilibrium charge separation — and hence the electrode potential — for that particular metal–ion couple.
Cell Potential and EMF
The potential difference between the two electrodes of a galvanic cell is the cell potential, measured in volts. It is the difference between the reduction potentials of the cathode and the anode. When no current is drawn from the cell, this potential difference is called the cell's electromotive force (EMF). The working relation NEET tests repeatedly is:
$$E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$$
NCERT pairs this with a representation convention: write the anode on the left, the cathode on the right. A single vertical line separates a metal from its electrolyte, and a double vertical line marks the salt bridge between the two electrolytes. Under this convention the EMF is the right-hand potential minus the left-hand potential:
$$E_{\text{cell}} = E_{\text{right}} - E_{\text{left}}$$
Take the silver–copper cell. The cell reaction $\ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)}$ splits into a reduction at silver and an oxidation at copper, so it is represented as $\ce{Cu(s) | Cu^2+(aq) || Ag+(aq) | Ag(s)}$, giving $E_{\text{cell}} = E_{\ce{Ag+ | Ag}} - E_{\ce{Cu^2+ | Cu}}$.
Once concentrations leave unity, EMF shifts. See how the Nernst equation corrects the standard cell potential.
Measuring Electrode Potential
A subtle but heavily examined point: the potential of an individual half-cell cannot be measured. Only the difference between two half-cell potentials — the cell EMF — is measurable. To assign absolute-looking numbers, one electrode is arbitrarily chosen as a reference and given zero potential.
That reference is the standard hydrogen electrode (SHE), represented $\ce{Pt(s) | H2(g) | H+(aq)}$, assigned a potential of exactly zero at all temperatures for the half-reaction $\ce{H+(aq) + e^- -> 1/2 H2(g)}$. It is a platinum electrode coated with platinum black, dipped in acid with $\ce{[H+]} = 1\ \text{M}$ while pure $\ce{H2}$ at 1 bar bubbles through. Any other electrode is then measured against it. Because $E^\circ_L = 0$ for SHE:
$$E^\circ_{\text{cell}} = E^\circ_R - E^\circ_L = E^\circ_R - 0 = E^\circ_R$$
So the EMF of the cell $\ce{Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) || Cu^2+(aq, 1 M) | Cu}$ is measured as 0.34 V, which is therefore the standard electrode potential of the $\ce{Cu^2+/Cu}$ couple. The same procedure with zinc gives $-0.76\ \text{V}$ for $\ce{Zn^2+/Zn}$.
Reconstruct the Daniell-cell EMF from the standard electrode potentials $E^\circ_{\ce{Cu^2+/Cu}} = +0.34\ \text{V}$ and $E^\circ_{\ce{Zn^2+/Zn}} = -0.76\ \text{V}$.
Reduction occurs at copper (cathode), oxidation at zinc (anode). Apply $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = \mathbf{1.10\ V}$. This matches the 1.1 V measured for the unit-concentration Daniell cell.
Sign Conventions and Spontaneity
The sign of $E^\circ$ encodes chemistry directly. A positive standard electrode potential means the species is reduced more readily than $\ce{H+}$, so its reduced form is more stable than hydrogen gas — copper's $+0.34\ \text{V}$ tells us $\ce{H+}$ cannot oxidise copper, which is why copper does not dissolve in HCl. A negative value, as for zinc's $-0.76\ \text{V}$, means $\ce{H+}$ can oxidise the metal; the metal is a strong reducing agent.
Reading down a standard-electrode-potential table, $E^\circ$ decreases: fluorine sits at the top ($+2.87\ \text{V}$, strongest oxidising agent) and lithium at the bottom (most powerful reducing agent). The single test of whether a galvanic cell works spontaneously is the sign of its computed $E^\circ_{\text{cell}}$.
| Half-reaction (reduction) | E° / V | Reading |
|---|---|---|
| $\ce{F2(g) + 2e^- -> 2F^-}$ | +2.87 | Strongest oxidising agent |
| $\ce{Ag+ + e^- -> Ag(s)}$ | +0.80 | Easily reduced |
| $\ce{Cu^2+ + 2e^- -> Cu(s)}$ | +0.34 | Reduced more readily than H⁺ |
| $\ce{Fe^2+ + 2e^- -> Fe(s)}$ | −0.44 | H⁺ can oxidise it |
| $\ce{Zn^2+ + 2e^- -> Zn(s)}$ | −0.76 | Strong reducing metal |
Do not flip the sign of E° when reversing the half-reaction inside the formula
The relation $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ already uses both values as reduction potentials. The subtraction handles the anode's reversal for you. Students who additionally negate the anode's $E^\circ$ (treating it as an oxidation potential and subtracting) double-count the sign and get the wrong EMF.
Always feed tabulated reduction potentials straight into cathode − anode; let the minus sign do the work.
EMF and Gibbs Energy
The electrical work a galvanic cell delivers links its EMF to the thermodynamics of the reaction. NCERT gives the relation:
$$\Delta_r G = -nFE_{\text{cell}}$$
where $n$ is the number of moles of electrons transferred and $F$ is the Faraday constant. A positive $E_{\text{cell}}$ forces $\Delta_r G$ negative, confirming the reaction is spontaneous — the very condition for a galvanic cell to function. A crucial distinction follows: $E_{\text{cell}}$ is an intensive property, independent of how much substance reacts, whereas $\Delta_r G$ is extensive because it carries the factor $n$. Multiplying the balanced equation by 2 doubles $\Delta_r G$ but leaves $E_{\text{cell}}$ unchanged — a favourite NEET assertion-reason hook.
Galvanic cells & electrode potential in one glance
- A galvanic (voltaic) cell converts the Gibbs energy of a spontaneous redox reaction into electrical energy; the Daniell cell gives 1.1 V at unit concentration.
- Each electrode is a half-cell: oxidation defines the anode (− in galvanic), reduction defines the cathode (+ in galvanic).
- Charge separation at the electrode–electrolyte interface produces the electrode potential; at unit concentration it is the standard electrode potential $E^\circ$ (a reduction potential by IUPAC).
- $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = E_{\text{right}} - E_{\text{left}}$, with anode written on the left.
- Only differences are measurable; the SHE $\ce{Pt(s)|H2(g)|H+(aq)}$ is the zero reference at all temperatures.
- A positive $E^\circ_{\text{cell}}$ means spontaneity; $\Delta_r G = -nFE_{\text{cell}}$ ties EMF to thermodynamics ($E$ intensive, $\Delta_r G$ extensive).