આ MCQ મોડ્યુલ આના પર આધારિત છે: Electrochemical Cells Electrode Potential
આ મૂલ્યાંકન આના પર આધારિત હશે: Electrochemical Cells Electrode Potential
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Every time you start a car, receive a pacemaker pulse, or swipe a phone off the charger, you are living off a controlled redox reaction. Electrochemistry is the branch of chemistry that studies the two-way conversion between chemical energy and electrical energy. In one direction, a spontaneous redox reaction is harnessed to push electrons through a wire (that is how a galvanic or voltaic cell works — a dry cell, a lead-acid battery, a lithium-ion pack). In the opposite direction, an electric current is used to drive a non-spontaneous reaction (an electrolytic cell — the route we take to extract aluminium, purify copper, or electroplate jewellery).
Electrochemistry underlies the entire modern energy landscape: portable batteries, rust prevention, hydrogen fuel cells, and the proposed Hydrogen Economy that aims to replace fossil fuels with hydrogen produced and consumed electrochemically. In this chapter we will build the quantitative tools — electrode potentials, the Nernst equation, conductance, Faraday's laws — that make such applications possible.
An electrochemical cell is any device in which chemical and electrical energy inter-convert. They come in two flavours:
| Feature | Galvanic (Voltaic) Cell | Electrolytic Cell |
|---|---|---|
| Energy direction | Chemical → Electrical | Electrical → Chemical |
| Reaction type | Spontaneous (ΔG < 0) | Non-spontaneous (requires ΔG > 0 input) |
| Anode polarity | Negative | Positive |
| Cathode polarity | Positive | Negative |
| Example | Daniell cell, dry cell, lead-acid battery | Electrolysis of NaCl, electroplating |
The classic Daniell cell pairs a zinc rod sitting in aqueous ZnSO₄ with a copper rod sitting in aqueous CuSO₄. The two solutions are linked by a salt bridge (a U-tube filled with a KCl–agar paste) and the two metals are joined through an external wire. The overall reaction that runs spontaneously is:
Separated into its two half-reactions:
As Zn²⁺ accumulates in the anode beaker, that solution would turn positive; as Cu²⁺ is consumed in the cathode beaker, it would turn negative. Either imbalance would instantly halt electron flow. The salt bridge carries Cl⁻ ions toward the anode and K⁺ ions toward the cathode, cancelling the build-up and keeping the circuit alive.
Chemists compress the full cell picture into a short line:
Left side = anode (oxidation), right side = cathode (reduction). A single bar | marks a phase boundary; the double bar || marks the salt bridge.
Every half-cell has an intrinsic tendency to gain electrons, called its reduction potential. This tendency cannot be measured in isolation — only the difference between two half-cells can be read on a voltmeter. To put every electrode on a common scale, chemists defined the Standard Hydrogen Electrode (SHE) as the zero point:
Every other electrode is coupled to the SHE under standard conditions (1 M ion concentration, 1 bar gas pressure, 298 K, pure solid) and the measured EMF is reported as E°, the standard electrode (reduction) potential.
| Half-reaction | E° (V) | Tendency |
|---|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.87 | Strongest oxidising agent |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | +1.51 | Strong oxidiser (KMnO₄) |
| Ag⁺ + e⁻ → Ag | +0.80 | Mild oxidiser |
| Cu²⁺ + 2e⁻ → Cu | +0.34 | Mild oxidiser |
| 2H⁺ + 2e⁻ → H₂ | 0.00 | Reference |
| Fe²⁺ + 2e⁻ → Fe | −0.44 | Mild reducer |
| Zn²⁺ + 2e⁻ → Zn | −0.76 | Strong reducer |
| Li⁺ + e⁻ → Li | −3.05 | Strongest reducing agent |
For the Daniell cell: \(E^{\circ}_{cell} = 0.34 - (-0.76) = +1.10\,\text{V}\). Because the result is positive, the reaction runs as written.
EMF is measured with a potentiometer rather than an ordinary voltmeter. A potentiometer draws essentially no current from the cell, so the reading reflects the true cell potential; a voltmeter would draw a small current, perturb the electrode concentrations and give a lower reading.
E° values apply only at standard conditions (1 M, 1 bar, 298 K). What happens when the concentrations are different? Walther Nernst (1889) supplied the answer: the electrode potential changes logarithmically with the activity ratio of products to reactants.
For a general reduction half-reaction Mn+ + n e⁻ → M(s):
Here R = 8.314 J K⁻¹ mol⁻¹, F = 96,500 C mol⁻¹, T = 298 K; the activity of a pure solid is taken as 1. Note: 0.0591 = 2.303 RT/F at 298 K.
where Q is the reaction quotient, written in the same product-over-reactant form you already know from Chapter 6 of Class 11.
As the cell discharges, Q grows; as Q → K (the equilibrium constant), Ecell → 0 and the cell can do no more work. Setting Ecell = 0:
A large positive E°cell therefore implies a huge K — the reaction runs essentially to completion.
The maximum electrical work a cell can deliver equals the drop in Gibbs energy of the redox reaction:
So three quantities — ΔG°, E°cell and K — are different faces of the same coin.
Given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V, find E°cell for Zn | Zn²⁺ || Cu²⁺ | Cu.
Answer: 1.10 V. The positive sign confirms the spontaneity of Zn + Cu²⁺ → Zn²⁺ + Cu.
For the Daniell cell (n = 2, E°cell = 1.10 V), compute Kc at 298 K.
Answer: Kc ≈ 1.7 × 10³⁷ — effectively complete reaction.
Calculate ΔG° (in kJ) for Zn + Cu²⁺ → Zn²⁺ + Cu at 298 K. Use F = 96,500 C mol⁻¹.
Answer: ΔG° ≈ −212.3 kJ mol⁻¹ — a strongly negative, clearly spontaneous process.
Find the EMF of Zn | Zn²⁺ (0.001 M) || Cu²⁺ (0.1 M) | Cu at 298 K.
n = 2. Reaction quotient: \(Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.001}{0.1} = 0.01\).
Answer: 1.16 V — slightly higher than standard because the "reactant" Cu²⁺ is more concentrated than the "product" Zn²⁺.
Find the EMF at 298 K of the cell Cu | Cu²⁺ (0.001 M) || Cu²⁺ (0.10 M) | Cu.
Both electrodes are Cu, so E°cell = 0. Only the concentration difference drives the cell. With n = 2 and Q = [dilute]/[concentrated] = 0.001/0.10 = 0.01:
Answer: 0.059 V. Ions spontaneously "flow" from the concentrated to the dilute half — nature's way of evening out concentrations.
Write the cell for Mg(s) + Cu²⁺ → Mg²⁺ + Cu(s) and compute E°cell. Given E°(Mg²⁺/Mg) = −2.37 V, E°(Cu²⁺/Cu) = +0.34 V.
Mg is oxidised (anode, more negative E°), Cu²⁺ is reduced (cathode).
Answer: 2.71 V — a much "hotter" cell than Zn/Cu.
For the half-cell 2H⁺(aq) + 2e⁻ → H₂(g, 1 bar) at pH = 3, find E at 298 K.
E° = 0. n = 2. Q = p(H₂)/[H⁺]² = 1/(10⁻³)² = 10⁶.
Answer: E ≈ −0.177 V. Every unit drop in [H⁺] lowers the hydrogen electrode potential by 0.0591 V — the principle behind the pH meter.
Enter the standard reduction potentials of any two electrodes. The tool identifies the anode (more negative E°) and cathode (more positive E°) and computes E°cell and ΔG°.
Q1. L1 Remember In a galvanic cell, the positive terminal is:
Q2. L3 Apply Write the overall reaction and compute E°cell for the cell above. (2 marks)
Q3. L3 Apply Use the Nernst equation to find Ecell at the given concentrations. (3 marks)
Q4. L3 Apply Compute ΔG° (in kJ) for the reaction. (2 marks)
Q5. L4 Analyse If the student replaces silver with copper (E° = +0.34 V) but keeps magnesium, will E°cell increase, decrease, or stay the same? Justify with a calculation. (3 marks)
Assertion (A): The EMF of a galvanic cell is measured using a potentiometer rather than a voltmeter.
Reason (R): A voltmeter draws some current from the cell, whereas a potentiometer measures the cell potential under essentially zero-current conditions.
Assertion (A): ΔG° of a galvanic cell can be calculated from its E°cell.
Reason (R): The maximum electrical work done by a reversible cell equals the decrease in Gibbs energy: ΔG° = −nFE°cell.
Assertion (A): E°cell of a concentration cell is zero.
Reason (R): Both electrodes of a concentration cell are of the same metal immersed in solutions of the same ion, so their standard potentials are identical.
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