Q.A Daniell (galvanic) cell is used as a power source to run a separate electrolytic cell. In the galvanic cell, a zinc plate dips in ZnSO4 solution (left half-cell) and a copper plate dips in CuSO4 solution (right half-cell), the two half-cells being joined by a salt bridge. This galvanic cell is wired to a lower electrolytic cell that has two electrodes, called electrode 'A' (on the left) and electrode 'B' (on the right), dipping in an electrolyte. The zinc plate of the galvanic cell is connected by a wire to electrode 'A', and the copper plate is connected by a wire to electrode 'B'. What will be the polarity (positive or negative) of electrodes 'A' and 'B' in the electrolytic cell?

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

The cell potential $E$ under non-standard conditions is given by:

Where:

• $E$ = cell potential under the given conditions (in volts)
• $E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
• $R$ = universal gas constant (8.314 J/mol·K)
• $T$ = temperature in Kelvin
• $n$ = number of moles of electrons transferred in the balanced half-reactions
• $F$ = Faraday constant (96,485 C/mol)
• $Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

• Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
• Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

So the Nernst equation becomes:

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V.

$E = 1.10 - \frac{0.0591}{2} \log_{10} \frac{0.1}{1.0}$ …