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Electrochemistry - Electrochemical cells

Grade 12ICSEChemistry

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

πŸ”‘Concepts

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An electrochemical cell consists of two electrodes (anode and cathode) in contact with an electrolyte. It is classified into Galvanic (Voltaic) cells, which convert chemical energy to electrical energy, and Electrolytic cells, which use electrical energy to drive non-spontaneous chemical reactions.

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In a Galvanic cell, oxidation occurs at the Anode (negative terminal) and reduction occurs at the Cathode (positive terminal). The mnemonic 'LOAN' (Left Oxidation Anode Negative) is helpful.

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The Salt Bridge is a U-shaped tube containing an inert electrolyte like KClKCl or KNO3KNO_3 in agar-agar. It completes the electrical circuit and maintains the electrical neutrality of the solutions in the two half-cells.

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The Standard Hydrogen Electrode (SHE) is the reference electrode used to measure electrode potentials. Its potential is arbitrarily taken as 0.00V0.00 V at all temperatures.

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

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Gibbs Free Energy (DeltaG\\Delta G): The electrical work done by a Galvanic cell is equal to the decrease in its Gibbs free energy. A reaction is spontaneous if Ξ”G<0\Delta G < 0, which implies Ecell>0E_{cell} > 0.

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The Nernst Equation relates the electrode potential or cell potential to the concentration of the species involved and the temperature.

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{2.303RT}{nF} \log Q

Ecell=Ecellβˆ˜βˆ’0.0591nlog⁑[Products][Reactants]Β atΒ 298KE_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \text{ at } 298 K

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

log⁑Kc=nEcell∘0.0591 at 298K\log K_c = \frac{n E^\circ_{cell}}{0.0591} \text{ at } 298 K

Q=IΓ—tΒ andΒ W=EeqΓ—IΓ—t96500Q = I \times t \text{ and } W = \frac{E_{eq} \times I \times t}{96500}

πŸ’‘Examples

Problem 1:

Calculate the EcellE_{cell} for the following cell at 298K298 K: Mg(s)∣Mg2+(0.001M)∣∣Cu2+(0.0001M)∣Cu(s)Mg(s) | Mg^{2+}(0.001 M) || Cu^{2+}(0.0001 M) | Cu(s). Given EMg2+/Mg∘=βˆ’2.36VE^\circ_{Mg^{2+}/Mg} = -2.36 V and ECu2+/Cu∘=+0.34VE^\circ_{Cu^{2+}/Cu} = +0.34 V.

Solution:

  1. Find Ecell∘E^\circ_{cell}: Ecell∘=Ecathodeβˆ˜βˆ’Eanode∘=0.34Vβˆ’(βˆ’2.36V)=2.70VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 V - (-2.36 V) = 2.70 V.
  2. Identify nn: The reaction is Mg+Cu2+β†’Mg2++CuMg + Cu^{2+} \rightarrow Mg^{2+} + Cu, so n=2n = 2.
  3. Apply Nernst Equation: Ecell=2.70βˆ’0.05912log⁑[Mg2+][Cu2+]E_{cell} = 2.70 - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}.
  4. Substitute values: Ecell=2.70βˆ’0.02955log⁑10βˆ’310βˆ’4=2.70βˆ’0.02955log⁑(10)=2.70βˆ’0.02955=2.67045VE_{cell} = 2.70 - 0.02955 \log \frac{10^{-3}}{10^{-4}} = 2.70 - 0.02955 \log(10) = 2.70 - 0.02955 = 2.67045 V.

Explanation:

We first determine the standard cell potential using reduction potentials. Then, the Nernst equation is used to adjust for non-standard concentrations. Since the concentration of Mg2+Mg^{2+} is 10 times that of Cu2+Cu^{2+}, the log term reduces the overall potential.

Problem 2:

Calculate the standard Gibbs energy change (DeltaG∘\\Delta G^\circ) for the reaction: 2Fe3+(aq)+2Iβˆ’(aq)β†’2Fe2+(aq)+I2(s)2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s), given Ecell∘=0.236VE^\circ_{cell} = 0.236 V.

Solution:

  1. Identify nn: The reaction involves the transfer of 22 electrons (2Iβˆ’β†’I2+2eβˆ’2I^- \rightarrow I_2 + 2e^-), so n=2n = 2.
  2. Use the formula: Ξ”G∘=βˆ’nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}.
  3. Substitute values: Ξ”G∘=βˆ’(2)Γ—(96500C/mol)Γ—(0.236V)\Delta G^\circ = -(2) \times (96500 C/mol) \times (0.236 V).
  4. Calculate: Ξ”G∘=βˆ’45548J/mol=βˆ’45.55kJ/mol\Delta G^\circ = -45548 J/mol = -45.55 kJ/mol.

Explanation:

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

Electrochemical cells - Revision Notes & Key Formulas | ICSE Class 12 Chemistry