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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 18.26

Complete and balance the following half–equations, and indicate whether oxidation or reduction is involved.

(a) ClO2(g) ClO3(aq) (acidic solution)

(b) MnO4(aq) MnO2(s) (acidic solution)

(c) SbH3(g) Sb(s) (basic solution)




Answer:

(a) ClO2(g) ClO3(aq) (acidic solution)

ClO2(g) + H2O(l) ClO3(aq)

ClO2(g) + H2O(l) ClO3(aq) + 2 H+(aq)

ClO2(g) + H2O(l) ClO3(aq) + 2 H+(aq) + e

One electron is required as a product to make both sides charge balanced.

Since an electron is lost, this is an oxidation (equivalently, the chlorine changes oxidation state from Cl+4 to Cl+5, a gain in oxidation number).



(b) MnO4(aq) MnO2(s) (acidic solution)

MnO4(aq) MnO2(s) + 2 H2O(l)

MnO4(aq) + 4 H+(aq) MnO2(s) + 2 H2O(l)

MnO4(aq) + 4 H+(aq) + 3 e MnO2(s) + 2 H2O(l)

Three electrons are required as a reactant to make both sides charge balanced.

Since an electron is gained, this is a reduction (equivalently, the manganese changes oxidation state from Mn+7 to Mn+4, a reduction in oxidation number).



(c) SbH3(g) Sb(s) (basic solution)

SbH3(g) Sb(s) + 3 H+(aq)

SbH3(g) Sb(s) + 3 H+(aq) + 3 e

SbH3(g) + 3 OH(aq) Sb(s) + 3 H+(aq) + 3 OH(aq) + 3 e

3 H+(aq) + 3 OH(aq) 3 H2O(l)

Adding the last two reactions gives:

SbH3(g) + 3 OH(aq) Sb(s) + 3 H2O(l) + 3 e

Three electrons are required as a product to make both sides charge balanced.

Since electrons are lost, this is an oxidation (equivalently, the antimony changes oxidation state from Sb–3 to Sb0, a gain in oxidation number).