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Redox Reactions - Redox Reactions and Electrode Processes

Grade 11CBSEChemistry

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

πŸ”‘Concepts

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An Electrode Process involves a redox reaction occurring at the interface between an electrode and an electrolyte. It consists of two half-reactions: oxidation at the anode and reduction at the cathode.

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A Galvanic (Voltaic) Cell is a device that converts chemical energy from a spontaneous redox reaction into electrical energy. A classic example is the Daniell Cell consisting of ZnZn and CuCu electrodes.

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Electrode Potential (EE) is the potential difference established between the metal electrode and the solution of its ions. When the concentrations of all species are unity (1 M1\, M) and the temperature is 298 K298\, K, it is called Standard Electrode Potential (E∘E^\circ).

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The Standard Hydrogen Electrode (SHE) serves as a reference electrode. It consists of platinum foil coated with platinum black in 1.0 M H+1.0\, M\, H^+ solution with H2H_2 gas bubbled at 1 bar1\, bar. Its potential is assigned as 0.00 V0.00\, V at all temperatures.

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The Electrochemical Series is the arrangement of various electrodes in the decreasing order of their standard reduction potentials. Substances with higher reduction potentials act as stronger oxidizing agents.

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

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

πŸ“Formulae

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

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

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

Ecell=Ecellβˆ˜βˆ’0.0591nlog⁑10Q(atΒ 298 K)E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log_{10} Q \quad \text{(at } 298\,K\text{)}

πŸ’‘Examples

Problem 1:

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

Solution:

In a Daniell cell, Copper acts as the cathode (reduction) and Zinc acts as the anode (oxidation). Using the formula: Ecell∘=Ecathodeβˆ˜βˆ’Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} Ecell∘=ECu2+/Cuβˆ˜βˆ’EZn2+/Zn∘E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Zn^{2+}/Zn} Ecell∘=0.34 Vβˆ’(βˆ’0.76 V)=+1.10 VE^\circ_{cell} = 0.34\, V - (-0.76\, V) = +1.10\, V

Explanation:

Since the cell potential is positive (+1.10 V+1.10\, V), the redox reaction is spontaneous in the forward direction.

Problem 2:

Represent the cell in which the following reaction takes place: Mg(s)+2Ag+(aq)β†’Mg2+(aq)+2Ag(s)Mg(s) + 2Ag^+(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s). Specify the anode and cathode.

Solution:

The cell notation is: Mg(s)∣Mg2+(aq)∣∣Ag+(aq)∣Ag(s)Mg(s) | Mg^{2+}(aq) || Ag^+(aq) | Ag(s) Anode: Mg(s)β†’Mg2+(aq)+2eβˆ’Mg(s) \rightarrow Mg^{2+}(aq) + 2e^- (Oxidation) \ Cathode: Ag+(aq)+eβˆ’β†’Ag(s)Ag^+(aq) + e^- \rightarrow Ag(s) (Reduction)

Explanation:

In cell notation, the anode (oxidation half-cell) is written on the left and the cathode (reduction half-cell) is written on the right, separated by a double vertical line representing the salt bridge.

Redox Reactions and Electrode Processes Revision - Class 11 Chemistry CBSE