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Redox Processes - Oxidation and reduction

Grade 11IBChemistry

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

πŸ”‘Concepts

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Oxidation is the loss of electrons, whereas reduction is the gain of electrons (OILΒ RIGOIL\ RIG).

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Oxidation state (or oxidation number) is the apparent charge an atom would have if all bonds were purely ionic. For example, in H2OH_2O, HH has an oxidation state of +1+1 and OO has βˆ’2-2.

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An oxidizing agent (oxidant) is the species that is reduced, as it accepts electrons from another species. A reducing agent (reductant) is the species that is oxidized, as it provides electrons.

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A disproportionation reaction is a specific type of redox reaction where the same element is simultaneously oxidized and reduced, such as 2H2O2(aq)β†’2H2O(l)+O2(g)2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g).

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In electrochemical cells, oxidation always occurs at the anode (ANOXANOX) and reduction always occurs at the cathode (REDΒ CATRED\ CAT).

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Voltaic (Galvanic) cells convert chemical energy into electrical energy via spontaneous redox reactions (EcellβŠ–>0E_{cell}^\ominus > 0).

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Electrolytic cells use electrical energy to drive non-spontaneous redox reactions (EcellβŠ–<0E_{cell}^\ominus < 0).

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The Activity Series ranks metals in order of their ability to displace one another from solutions. A metal higher in the series is a stronger reducing agent.

πŸ“Formulae

EcellβŠ–=EcathodeβŠ–βˆ’EanodeβŠ–E_{cell}^\ominus = E_{cathode}^\ominus - E_{anode}^\ominus

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

Oxidation:Β Mβ†’Mn++neβˆ’Oxidation:\ M \rightarrow M^{n+} + ne^-

Reduction:Β Xn++neβˆ’β†’XReduction:\ X^{n+} + ne^- \rightarrow X

πŸ’‘Examples

Problem 1:

Determine the oxidation state of sulfur in the thiosulfate ion, S2O32βˆ’S_2O_3^{2-}.

Solution:

Let the oxidation state of SS be xx. The oxidation state of OO is usually βˆ’2-2. The sum of oxidation states must equal the charge of the ion: 2(x)+3(βˆ’2)=βˆ’22(x) + 3(-2) = -2. Therefore, 2xβˆ’6=βˆ’2β‡’2x=+4β‡’x=+22x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2.

Explanation:

By applying the rule that the sum of oxidation numbers equals the net charge of the polyatomic ion, we calculate the average oxidation state of sulfur to be +2+2.

Problem 2:

Calculate the standard cell potential (EcellβŠ–E_{cell}^\ominus) for a voltaic cell consisting of a Zinc electrode in Zn2+(aq)Zn^{2+}(aq) and a Copper electrode in Cu2+(aq)Cu^{2+}(aq) given the standard reduction potentials: EβŠ–(Zn2+/Zn)=βˆ’0.76Β VE^\ominus(Zn^{2+}/Zn) = -0.76\ V and EβŠ–(Cu2+/Cu)=+0.34Β VE^\ominus(Cu^{2+}/Cu) = +0.34\ V.

Solution:

The more positive potential is the cathode (reduction). Cathode: Cu2++2eβˆ’β†’CuCu^{2+} + 2e^- \rightarrow Cu. Anode: Znβ†’Zn2++2eβˆ’Zn \rightarrow Zn^{2+} + 2e^-. EcellβŠ–=EcathodeβŠ–βˆ’EanodeβŠ–=(+0.34Β V)βˆ’(βˆ’0.76Β V)=+1.10Β VE_{cell}^\ominus = E_{cathode}^\ominus - E_{anode}^\ominus = (+0.34\ V) - (-0.76\ V) = +1.10\ V

Explanation:

In a voltaic cell, the half-cell with the higher reduction potential acts as the cathode. The standard cell potential is the difference between the reduction potential of the cathode and the reduction potential of the anode.

Problem 3:

Balance the following half-reaction in acidic solution: MnO4βˆ’(aq)β†’Mn2+(aq)MnO_4^-(aq) \rightarrow Mn^{2+}(aq).

Solution:

MnO4βˆ’(aq)+8H+(aq)+5eβˆ’β†’Mn2+(aq)+4H2O(l)MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l)

Explanation:

  1. Balance atoms other than HH and OO (MnMn is balanced). 2. Balance OO by adding H2OH_2O. 3. Balance HH by adding H+H^+. 4. Balance the overall charge by adding electrons (eβˆ’e^-).
Oxidation and reduction - Revision Notes & Key Formulas | IB Grade 11 Chemistry