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Electrochemistry - Electrochemical Cells

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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A Galvanic (Voltaic) cell converts chemical energy from a spontaneous redox reaction into electrical energy. It consists of two half-cells connected by a salt bridge. The anode is the electrode where oxidation occurs (negative polarity), and the cathode is where reduction occurs (positive polarity).

A standard Daniel cell showing a zinc electrode in zinc sulfate solution and a copper electrode in copper sulfate solution.
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The Salt Bridge completes the electrical circuit and maintains electrical neutrality in the half-cells by allowing the migration of ions. It typically contains an inert electrolyte like KClKCl or KNO3KNO_3 in agar-agar gel.

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Electrode Potential is the potential difference established between the metal electrode and its ion solution. Under standard conditions (1 M1\text{ M} concentration, 298 K298\text{ K}), it is called Standard Electrode Potential (E∘E^\circ).

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The Cell EMF (EcellE_{cell}) is the potential difference between the two electrodes of the cell when no current is drawn. It is calculated as Ecell=Ecathodeβˆ’EanodeE_{cell} = E_{cathode} - E_{anode} using reduction potentials.

πŸ“Formulae

Ecell∘=Ecathodeβˆ˜βˆ’Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Ecell=Ecellβˆ˜βˆ’2.303RTnFlog⁑QE_{cell} = E^\circ_{cell} - \frac{2.303 RT}{nF} \log Q

Ecell=Ecellβˆ˜βˆ’0.0591nlog⁑[AnodeΒ Ion][CathodeΒ Ion]Β (atΒ 298Β K)E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[\text{Anode Ion}]}{[\text{Cathode Ion}]} \text{ (at 298 K)}

Ξ”G∘=βˆ’nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}

log⁑Kc=nEcell∘0.0591\log K_c = \frac{n E^\circ_{cell}}{0.0591}

πŸ’‘Examples

Problem 1:

Calculate the emf of the following cell at 298Β K298\text{ K}: Mg(s)∣Mg2+(0.001Β M)∣∣Cu2+(0.0001Β M)∣Cu(s)Mg(s) | Mg^{2+}(0.001\text{ M}) || Cu^{2+}(0.0001\text{ M}) | Cu(s). Given EMg2+/Mg∘=βˆ’2.37Β VE^\circ_{Mg^{2+}/Mg} = -2.37\text{ V} and ECu2+/Cu∘=+0.34Β VE^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}.

Solution:

  1. Calculate Ecell∘E^\circ_{cell}: Ecell∘=Ecathodeβˆ˜βˆ’Eanode∘=0.34βˆ’(βˆ’2.37)=2.71Β VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-2.37) = 2.71\text{ V}.
  2. Identify nn: For Mgβ†’Mg2++2eβˆ’Mg \rightarrow Mg^{2+} + 2e^- and Cu2++2eβˆ’β†’CuCu^{2+} + 2e^- \rightarrow Cu, n=2n = 2.
  3. Apply Nernst Equation: Ecell=2.71βˆ’0.05912log⁑[Mg2+][Cu2+]E_{cell} = 2.71 - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}.
  4. Substitute values: Ecell=2.71βˆ’0.02955log⁑10βˆ’310βˆ’4=2.71βˆ’0.02955log⁑(10)E_{cell} = 2.71 - 0.02955 \log \frac{10^{-3}}{10^{-4}} = 2.71 - 0.02955 \log(10).
  5. Ecell=2.71βˆ’0.02955(1)=2.68045Β VE_{cell} = 2.71 - 0.02955(1) = 2.68045\text{ V}.

Explanation:

The standard cell potential is calculated first. Since the concentrations are not 1Β M1\text{ M}, the Nernst equation is used to find the actual cell potential. The reaction quotient QQ is the ratio of product ion concentration to reactant ion concentration.

Problem 2:

Calculate the standard Gibbs energy Ξ”G∘\Delta G^\circ for the reaction: Zn(s)+Cu2+(aq)β†’Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s). Given Ecell∘=1.1Β VE^\circ_{cell} = 1.1\text{ V} and F=96500Β C/molF = 96500\text{ C/mol}.

Solution:

  1. Identify nn: In this redox reaction, 22 electrons are transferred, so n=2n = 2.
  2. Use the formula: Ξ”G∘=βˆ’nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}.
  3. Ξ”G∘=βˆ’(2)Γ—(96500Β C/mol)Γ—(1.1Β V)\Delta G^\circ = -(2) \times (96500\text{ C/mol}) \times (1.1\text{ V}).
  4. Ξ”G∘=βˆ’212300Β J/mol=βˆ’212.3Β kJ/mol\Delta G^\circ = -212300\text{ J/mol} = -212.3\text{ kJ/mol}.

Explanation:

Standard Gibbs energy is directly proportional to the standard cell potential. The negative sign indicates that the reaction is thermodynamically spontaneous under standard conditions.

Problem 3:

Calculate the equilibrium constant (KcK_c) for the reaction occurring in a Daniel cell at 298Β K298\text{ K} if the standard cell potential is 1.1Β V1.1\text{ V}. The reaction is: Zn(s)+Cu2+(aq)β†’Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s) Use the relation log⁑Kc=nEcell∘0.0591\log K_c = \frac{n E^\circ_{cell}}{0.0591}.

Schematic of a Daniel cell used to calculate equilibrium constant.

Solution:

  1. Identify the number of electrons transferred: n=2n = 2 for the Zn/CuZn/Cu system.
  2. Use the Nernst-derived formula for equilibrium: log⁑Kc=2Γ—1.10.0591\log K_c = \frac{2 \times 1.1}{0.0591}
  3. Calculate the value: log⁑Kc=2.20.0591β‰ˆ37.225\log K_c = \frac{2.2}{0.0591} \approx 37.225
  4. Find the antilog: Kc=1037.225β‰ˆ2Γ—1037K_c = 10^{37.225} \approx 2 \times 10^{37}

Explanation:

At equilibrium, Ecell=0E_{cell} = 0, which allows us to relate the standard cell potential directly to the equilibrium constant of the redox reaction.

Problem 4:

A silver-copper cell is constructed. Given EAg+/Ag∘=+0.80 VE^\circ_{Ag^+/Ag} = +0.80\text{ V} and ECu2+/Cu∘=+0.34 VE^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}. Write the cell notation and calculate the standard EMF of the cell.

Galvanic cell consisting of a copper anode and a silver cathode.

Solution:

  1. Compare reduction potentials: EAg+/Ag∘>ECu2+/Cu∘E^\circ_{Ag^+/Ag} > E^\circ_{Cu^{2+}/Cu}, so Silver acts as the cathode and Copper as the anode.
  2. Write the cell representation: Cu(s)∣Cu2+(aq)∣∣Ag+(aq)∣Ag(s)Cu(s) | Cu^{2+}(aq) || Ag^+(aq) | Ag(s)
  3. Calculate Ecell∘E^\circ_{cell}: Ecell∘=Ecathodeβˆ˜βˆ’Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} Ecell∘=0.80Β Vβˆ’0.34Β V=0.46Β VE^\circ_{cell} = 0.80\text{ V} - 0.34\text{ V} = 0.46\text{ V}
  4. Final result: The standard EMF is 0.46Β V0.46\text{ V}.

Explanation:

In a galvanic cell, the electrode with the higher reduction potential undergoes reduction and is designated as the cathode.

Electrochemical Cells Class 12 Notes & Examples