Balance the following full redox reactions using the lowest integer coefficients

Question

Balance the following full redox reactions using the lowest integer coefficients, in basic conditions.

NiO2(s) + Zn(s) ? Ni(OH)2(s) + ZnO22?(aq)

Explanation / Answer

NiO2(s)====>Ni(OH)2(s) Need two H on the left so we add H+ 2H+(aq) + NiO2(s)====>Ni(OH)2(s) needs two electrons on the left to balance charge 2e- + 2H+(aq) + NiO2(s)====>Ni(OH)2(s) Zn(s)====>Zn(OH)2(s) Needs two O on the left so we add water. 2H2O(l) + Zn(s)====>Zn(OH)2(s) Needs two H on the left so we add H+. 2H2O(l) + Zn(s)====>Zn(OH)2(s) + 2H+(aq) needs two electrons on the right to balance charge 2H2O(l) + Zn(s)====>Zn(OH)2(s) + 2H+(aq) + 2e- Now we have both half reactions balanced. 2e- + 2H+(aq) + NiO2(s)====>Ni(OH)2(s) 2H2O(l) + Zn(s)====>Zn(OH)2(s) + 2H+(aq) + 2e- combine them to get: NiO2(s) + 2H2O(l) + Zn(s)====>Ni(OH)2(s) + Zn(OH)2(s) The 2e- and 2H+ cancel each other out because they're on both sides of the equation so we just get rid of them

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