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Redox Processes - Electrochemical cells

Grade 11IBChemistry

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

πŸ”‘Concepts

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An electrochemical cell consists of two half-cells connected by an external circuit and, in the case of voltaic cells, a salt bridge.

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Oxidation always occurs at the anode, and reduction always occurs at the cathode (Mnemonic: AN OX and RED CAT).

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Voltaic (Galvanic) Cells convert chemical energy from spontaneous redox reactions into electrical energy. The anode is the negative electrode (EβŠ–E^{\ominus} is more negative), and the cathode is the positive electrode (EβŠ–E^{\ominus} is more positive).

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Electrolytic Cells use electrical energy to drive non-spontaneous redox reactions. The anode is connected to the positive terminal of the power supply, and the cathode is connected to the negative terminal.

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The Salt Bridge in a voltaic cell completes the circuit and maintains electrical neutrality by allowing the migration of ions: anions move towards the anode and cations move towards the cathode.

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The Standard Hydrogen Electrode (SHE) is used as a reference point to measure standard electrode potentials (EβŠ–E^{\ominus}), with a defined value of 0.00 V0.00\,V at 298 K298\,K, 100 kPa100\,kPa, and 1.0 mol dmβˆ’31.0\,mol\,dm^{-3} H+(aq)H^{+}(aq).

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Standard cell potential (EcellβŠ–E_{cell}^{\ominus}) can be used to predict the spontaneity of a reaction. A positive EcellβŠ–E_{cell}^{\ominus} indicates a spontaneous reaction (Ξ”GβŠ–<0\Delta G^{\ominus} < 0).

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Cell notation is a shorthand representation: Anode(s)∣Anode Ion(aq)∣∣Cathode Ion(aq)∣Cathode(s)Anode(s) | Anode\,Ion(aq) || Cathode\,Ion(aq) | Cathode(s). For example: Zn(s)∣Zn2+(aq)∣∣Cu2+(aq)∣Cu(s)Zn(s) | Zn^{2+}(aq) || Cu^{2+}(aq) | Cu(s).

πŸ“Formulae

EcellβŠ–=EreductionβŠ–βˆ’EoxidationβŠ–E_{cell}^{\ominus} = E_{reduction}^{\ominus} - E_{oxidation}^{\ominus}

Ξ”GβŠ–=βˆ’nFEcellβŠ–\Delta G^{\ominus} = -nFE_{cell}^{\ominus}

Q=ItQ = It

n=QzFn = \frac{Q}{zF}

πŸ’‘Examples

Problem 1:

Calculate the standard cell potential (EcellβŠ–E_{cell}^{\ominus}) for a voltaic cell based on the following half-reactions: Zn2+(aq)+2eβˆ’β†’Zn(s)Zn^{2+}(aq) + 2e^{-} \rightarrow Zn(s) (EβŠ–=βˆ’0.76 VE^{\ominus} = -0.76\,V) and Cu2+(aq)+2eβˆ’β†’Cu(s)Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s) (EβŠ–=+0.34 VE^{\ominus} = +0.34\,V). Determine if the reaction is spontaneous.

Solution:

  1. Identify the cathode and anode: The more positive value is the cathode (Cu2+/CuCu^{2+}/Cu), and the more negative value is the anode (Zn2+/ZnZn^{2+}/Zn).
  2. Use the formula: EcellβŠ–=EcathodeβŠ–βˆ’EanodeβŠ–=0.34 Vβˆ’(βˆ’0.76 V)=+1.10 VE_{cell}^{\ominus} = E_{cathode}^{\ominus} - E_{anode}^{\ominus} = 0.34\,V - (-0.76\,V) = +1.10\,V.
  3. Since EcellβŠ–>0E_{cell}^{\ominus} > 0, the reaction is spontaneous.

Explanation:

In a voltaic cell, electrons flow from the species with the more negative electrode potential (reducing agent) to the species with the more positive electrode potential (oxidizing agent). The positive cell potential confirms a negative Gibbs free energy change.

Problem 2:

Predict the products formed at the electrodes during the electrolysis of molten sodium chloride (NaCl(l)NaCl(l)).

Solution:

At the Negative Electrode (Cathode): Na+(l)+eβˆ’β†’Na(l)Na^{+}(l) + e^{-} \rightarrow Na(l) At the Positive Electrode (Anode): 2Clβˆ’(l)β†’Cl2(g)+2eβˆ’2Cl^{-}(l) \rightarrow Cl_{2}(g) + 2e^{-}

Explanation:

In molten salts, only the constituent ions are present. Sodium ions are reduced to sodium metal at the cathode, and chloride ions are oxidized to chlorine gas at the anode.

Electrochemical cells - Revision Notes & Key Formulas | IB Grade 11 Chemistry