krit.club logo

Redox Reactions - Electrode Processes

Grade 11ICSEChemistry

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

🔑Concepts

•

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 (AN OXAN\ OX and RED CATRED\ CAT).

•

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

•

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.

•

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).

•

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.

•

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.

•

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

📐Formulae

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

Ecell∘=Ereduction∘−Eoxidation∘E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation}

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

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

1 F≈96500 C mol−11\ F \approx 96500\ C\ mol^{-1}

💡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∘=−0.76 VE^\circ_{Zn^{2+}/Zn} = -0.76\ V and ECu2+/Cu∘=+0.34 VE^\circ_{Cu^{2+}/Cu} = +0.34\ V.

Solution:

Ecell∘=ECu2+/Cu∘−EZn2+/Zn∘=0.34 V−(−0.76 V)=+1.10 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 Br−(aq)Br^-(aq) under standard conditions. Given EFe3+/Fe2+∘=0.77 VE^\circ_{Fe^{3+}/Fe^{2+}} = 0.77\ V and EBr2/Br−∘=1.09 VE^\circ_{Br_2/Br^-} = 1.09\ V.

Solution:

The cell reaction would be 2Fe3++2Br−→2Fe2++Br22Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2. Here, Fe3+Fe^{3+} is reduced (cathode) and Br−Br^- is oxidized (anode). Ecell∘=EFe3+/Fe2+∘−EBr2/Br−∘=0.77 V−1.09 V=−0.32 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 Br−Br^-.