1. Reverse one of the half-rxns into an oxidation step such that the sum of the two 1/2 reactions E° give a positive E° cll

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To write Spontaneous Redox Reactions:
1. Reverse one of the half-rxns into an oxidation step such that the sum of the two 1/2 reactions E°
give a positive E cell
2. Add the two half-rxns to obtained an overall redox reaction in such a way that the number of electrons cancel
from both sides of the equation.
*NOTE: Changing the "balancing" coefficients of a half-rxn does NOT change the E° value. An electrode potential, E
is an intensive property, one that does not depend on the amount of substance present. The potential is the ratio
of energy to charge, when the coefficients are changed (increasing the amount of substance) both the energy and
charge increases proportionally and therefore: the energy to charge ratio does not change: The potential, E°, does
not change.
Examples:
1. Hg,Cl2 + 2 e 5 2 Hg + 2 CI
E° = + 0.27
PBSO, + 2 e 5 Pb + So,?
E° = + 0.36
2. A3* + 3 e 5
Al
E = - 1.67
Br2 + 2 e
5 2 Br
E° = + 1.09
Transcribed Image Text:To write Spontaneous Redox Reactions: 1. Reverse one of the half-rxns into an oxidation step such that the sum of the two 1/2 reactions E° give a positive E cell 2. Add the two half-rxns to obtained an overall redox reaction in such a way that the number of electrons cancel from both sides of the equation. *NOTE: Changing the "balancing" coefficients of a half-rxn does NOT change the E° value. An electrode potential, E is an intensive property, one that does not depend on the amount of substance present. The potential is the ratio of energy to charge, when the coefficients are changed (increasing the amount of substance) both the energy and charge increases proportionally and therefore: the energy to charge ratio does not change: The potential, E°, does not change. Examples: 1. Hg,Cl2 + 2 e 5 2 Hg + 2 CI E° = + 0.27 PBSO, + 2 e 5 Pb + So,? E° = + 0.36 2. A3* + 3 e 5 Al E = - 1.67 Br2 + 2 e 5 2 Br E° = + 1.09
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