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Redox Reactions - Electrode Processes

Grade 11ICSEChemistry

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

πŸ”‘Concepts

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Electrode processes involve the transfer of electrons at the interface between an electrolyte and an electrode. Oxidation occurs at the Anode, while reduction occurs at the Cathode (ANΒ OXAN\ OX and REDΒ CATRED\ CAT).

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In a Galvanic (Voltaic) Cell, chemical energy is converted into electrical energy through spontaneous redox reactions. The anode is negative (βˆ’-) and the cathode is positive (++).

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The Standard Electrode Potential (E∘E^\circ) is the potential developed when a metal electrode is in contact with a 1 M1\ M solution of its own ions at 298 K298\ K and 1 bar1\ bar pressure.

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The Standard Hydrogen Electrode (SHE) serves as the reference electrode with an assigned potential of 0.00 V0.00\ V. It is represented as Pt(s)∣H2(g,1 bar)∣H+(aq,1 M)Pt(s) | H_2(g, 1\ bar) | H^+(aq, 1\ M).

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The Electrochemical Series is an arrangement of electrodes in increasing order of their standard reduction potentials. A lower (more negative) E∘E^\circ indicates a stronger reducing agent, while a higher (more positive) E∘E^\circ indicates a stronger oxidizing agent.

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A Salt Bridge is a UU-shaped tube containing an inert electrolyte like KClKCl or KNO3KNO_3 in agar-agar. It completes the circuit and maintains electrical neutrality in the half-cells by allowing ion migration.

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The feasibility of a redox reaction can be predicted using Ecell∘E^\circ_{cell}. If Ecell∘>0E^\circ_{cell} > 0, the reaction is spontaneous.

πŸ“Formulae

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

Ecell∘=Ereductionβˆ˜βˆ’Eoxidation∘E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation}

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

Wmax=nFEcell∘W_{max} = nFE^\circ_{cell}

1Β Fβ‰ˆ96500Β CΒ molβˆ’11\ F \approx 96500\ C\ mol^{-1}

πŸ’‘Examples

Problem 1:

Calculate the standard EMF (Ecell∘E^\circ_{cell}) of the cell represented as: Zn(s)∣Zn2+(aq)∣∣Cu2+(aq)∣Cu(s)Zn(s) | Zn^{2+}(aq) || Cu^{2+}(aq) | Cu(s). Given EZn2+/Zn∘=βˆ’0.76Β VE^\circ_{Zn^{2+}/Zn} = -0.76\ V and ECu2+/Cu∘=+0.34Β VE^\circ_{Cu^{2+}/Cu} = +0.34\ V.

Solution:

Ecell∘=ECu2+/Cuβˆ˜βˆ’EZn2+/Zn∘=0.34Β Vβˆ’(βˆ’0.76Β V)=+1.10Β VE^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Zn^{2+}/Zn} = 0.34\ V - (-0.76\ V) = +1.10\ V.

Explanation:

The electrode with the higher reduction potential acts as the cathode (CuCu), and the one with the lower potential acts as the anode (ZnZn).

Problem 2:

Predict if Fe3+(aq)Fe^{3+}(aq) can oxidize Brβˆ’(aq)Br^-(aq) under standard conditions. Given EFe3+/Fe2+∘=0.77Β VE^\circ_{Fe^{3+}/Fe^{2+}} = 0.77\ V and EBr2/Brβˆ’βˆ˜=1.09Β VE^\circ_{Br_2/Br^-} = 1.09\ V.

Solution:

The cell reaction would be 2Fe3++2Brβˆ’β†’2Fe2++Br22Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2. Here, Fe3+Fe^{3+} is reduced (cathode) and Brβˆ’Br^- is oxidized (anode). Ecell∘=EFe3+/Fe2+βˆ˜βˆ’EBr2/Brβˆ’βˆ˜=0.77Β Vβˆ’1.09Β V=βˆ’0.32Β VE^\circ_{cell} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Br_2/Br^-} = 0.77\ V - 1.09\ V = -0.32\ V.

Explanation:

Since the Ecell∘E^\circ_{cell} is negative, the reaction is non-spontaneous. Therefore, Fe3+Fe^{3+} cannot oxidize Brβˆ’Br^-.

Electrode Processes - Revision Notes & Key Formulas | ICSE Class 11 Chemistry