Question

Balance the following in basic solution. (Omit states-of-matter from your answer. Use the lowest possible whole number coefficients.) (a) Al(s) + Cr2O72?(aq) ? Al(OH)3(s) + Cr(OH)4?(aq) (state: Oxidation reaction/ Reduction reaction/ Net equation) (b) ClO3?(aq) + Cu(OH)2(s) ? ClO4?(aq) + CuOH(s) (state: Oxidation reaction/ Reduction reaction/ Net equation) (c) NO3?(aq) + H2(g) ? NO(g) (state: Oxidation reaction/ Reduction reaction/ Net equation)

Answer 1

Al(s) + Cr₂O₇^2-(aq) ? Al(OH)₃(s) + Cr(OH)_4?(aq)

get the oxidation states

Al ==== Al (OH)₃

0............+3....-3

Al goes from 0 to +3 (this is a oxidation)

Balance the hydrogens

Al + 3H₂O ==== Al (OH)₃ + 3 H⁺ (balance like if you had acidic conditions)

balance the electrons

Al + 3H₂O ==== Al (OH)₃ + 3 H⁺ + 3e^-

Add OH on each side of the equations:

Al + 3H₂O + 3 OH ==== Al (OH)₃ + 3 H⁺ + 3 OH + 3e^-

Al + 3H₂O + 3 OH ==== Al (OH)₃ + 3 H₂O  + 3e^-

Al + 3 OH ==== Al (OH)₃ +  3e^-, simplify the terms

Cr₂O₇^2-(aq) = Cr(OH)₄^-(aq)

Cr goes from +6 to +3 (this is an reduction)

Balance the number of Cr, multiply by 2 the right side to get:

Cr₂O₇^2-(aq) = 2 Cr(OH)₄^-(aq)

add water to balance the oxygens

Cr₂O₇^2-(aq) + H₂O = 2 Cr(OH)₄^-(aq), balance the hydrogens

Cr₂O₇^2-(aq) + H₂O + 6H⁺ = 2 Cr(OH)₄^-(aq), balance the electrons

Cr₂O₇^2-(aq) + H₂O + 6H⁺ + 6e^- = 2 Cr(OH)₄^-(aq), add OH on both sides of the equation

Cr₂O₇^2-(aq) + H₂O + 6H⁺ + 6e^- + 6 OH = 2 Cr(OH)₄^-(aq) + 6 OH

Cr₂O₇^2-(aq)+ 7 H₂O + 6e^- = 2 Cr(OH)₄^-(aq) + 6 OH^-

add the equations

Cr₂O₇^2-(aq) + 7 H₂O + 6e^- = 2 Cr(OH)₄^-(aq) + 6 OH^-

Al + 3 OH ==== Al (OH)₃ +  3e^- x (2) , multiply by 2 to match the number of electrons of the first equation

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Cr₂O₇^2-(aq) + 7 H₂O + 6e^- = 2 Cr(OH)₄^-(aq) + 6 OH^-

2 Al + 6 OH ==== 2 Al (OH)₃ + 6 e^-

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Cr₂O₇^2-(aq) + 7 H₂O + 6e^- + 2 Al + 6 OH = 2 Cr(OH)₄^-(aq) + 6 OH^- + 2 Al (OH)₃ + 6 e^-

remove similar terms

Cr₂O₇^2-(aq) + 7 H₂O (l)  + 2 Al(s) = 2 Cr(OH)₄^-(aq) + 2 Al (OH)₃ (s)

B)

ClO3^-(aq) + Cu(OH)2(s) = ClO4^-(aq) + CuOH(s)

ClO₃^-(aq) = ClO₄^-,

Cl goes from +5 to +7 , this is an oxidation

Cu(OH)₂ = CuOH

Cu goes from +2 to +1, this is a reduction

ClO₃^-(aq) = ClO₄^-(aq), add water

ClO₃^-(aq) + H₂O = ClO₄^-(aq)

ClO₃^-(aq) + H₂O = ClO₄^-(aq) + 2H⁺ + 2e^- , add OH^-

ClO₃^-(aq) + H₂O + 2 OH = ClO₄^-(aq) + 2H⁺ + 2e^- + 2 OH^-

ClO₃^-(aq) + 2 OH^- = ClO₄^-(aq) + 2e^- + H₂O

The other reaction:

Cu(OH)₂(s) = CuOH(s)

add water

Cu(OH)₂(s) + H⁺ + e^- = CuOH(s) + H₂O, add OH on both sides

Cu(OH)₂(s) + OH + H⁺ + e^- = CuOH(s) + H₂O + OH

Cu(OH)₂(s) + H₂O + e^- = CuOH(s) + H₂O + OH, remove similar terms

Cu(OH)₂(s) + e^- = CuOH(s) + OH

Add the equations

Cu(OH)₂(s)  + e^- = CuOH(s)  + OH x (2) , multiply this by the number of electrons in the second equation

ClO₃^-(aq)   + 2 OH^- ? ClO₄^-(aq) + 2e^- + H₂O

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2 Cu(OH)₂(s)  + 2 e^- = 2 CuOH(s)  + 2 OH^-  

ClO₃^-(aq)   + 2 OH^- = ClO₄^-(aq) + 2e^- + H₂O

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2 Cu(OH)₂(s)  + ClO₃^-(aq) = 2 CuOH(s) + ClO₄^-(aq) + H₂O

*hope this answer is helpful =)