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Electrochemistry - Galvanic cells

Grade 12ICSEChemistry

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

🔑Concepts

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A Galvanic (or Voltaic) cell is an electrochemical cell that converts chemical energy from a spontaneous redox reaction into electrical energy.

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In a Galvanic cell, the Anode is the negative electrode where oxidation occurs (LossLoss of electrons), and the Cathode is the positive electrode where reduction occurs (GainGain of electrons). A common mnemonic is LOAN: Left, Oxidation, Anode, Negative.

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The Salt Bridge contains an inert electrolyte like KClKCl, KNO3KNO_3, or NH4NO3NH_4NO_3 in agar-agar gel. It maintains electrical neutrality in both half-cells and completes the internal circuit by allowing ion flow.

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Cell Representation: The anode is written on the left and the cathode on the right. For example: Zn(s)∣Zn2+(aq)∣∣Cu2+(aq)∣Cu(s)Zn(s) | Zn^{2+}(aq) || Cu^{2+}(aq) | Cu(s), where '∣|' represents a phase boundary and '∣∣||' represents the salt bridge.

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Standard Electrode Potential (E∘E^\circ): The potential of an electrode measured under standard conditions (1 M1\,M concentration, 298 K298\,K temperature, and 1 bar1\,bar pressure) relative to the Standard Hydrogen Electrode (SHESHE), which is assigned a value of 0.00 V0.00\,V.

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Electromotive Force (EMF): The potential difference between the two electrodes when no current is drawn from the cell. Ecell=Ecathode−EanodeE_{cell} = E_{cathode} - E_{anode} (using reduction potentials).

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Spontaneity: For a reaction to be spontaneous in a Galvanic cell, the Gibbs Free Energy change must be negative (ΔG<0\Delta G < 0), which implies the cell potential must be positive (Ecell>0E_{cell} > 0).

📐Formulae

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Ecell=Ecell∘−2.303RTnFlog⁡QE_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log Q

Ecell=Ecell∘−0.0591nlog⁡[Anode ion][Cathode ion] (at 298 K)E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[Anode\,ion]}{[Cathode\,ion]} \text{ (at } 298\,K\text{)}

ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}

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

💡Examples

Problem 1:

Calculate the EMF of the following cell at 298 K298\,K: Mg(s)∣Mg2+(0.1 M)∣∣Cu2+(0.01 M)∣Cu(s)Mg(s) | Mg^{2+}(0.1\,M) || Cu^{2+}(0.01\,M) | Cu(s). Given EMg2+/Mg∘=−2.37 VE^\circ_{Mg^{2+}/Mg} = -2.37\,V and ECu2+/Cu∘=+0.34 VE^\circ_{Cu^{2+}/Cu} = +0.34\,V.

Solution:

  1. Calculate Ecell∘E^\circ_{cell}: Ecell∘=Ecathode∘−Eanode∘=0.34 V−(−2.37 V)=2.71 VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34\,V - (-2.37\,V) = 2.71\,V.
  2. Identify nn: In the reaction Mg+Cu2+→Mg2++CuMg + Cu^{2+} \rightarrow Mg^{2+} + Cu, the number of electrons transferred is n=2n = 2.
  3. Apply Nernst Equation: Ecell=Ecell∘−0.05912log⁡[Mg2+][Cu2+]E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}.
  4. Substitute values: Ecell=2.71−0.02955log⁡0.10.01=2.71−0.02955log⁡(10)=2.71−0.02955=2.68045 VE_{cell} = 2.71 - 0.02955 \log \frac{0.1}{0.01} = 2.71 - 0.02955 \log(10) = 2.71 - 0.02955 = 2.68045\,V.

Explanation:

First, the standard cell potential is determined. Then, the Nernst equation is used to account for the non-standard concentrations of the ions. Since the concentration of the anode ion is higher than the cathode ion, the cell potential decreases slightly from the standard value.