Answer:
See explanation...
Explanation:
To ID half-rxns as oxidation or reduction, look where the electrons are posted in the equation... That is,
if the electrons are on the product side of the equation => oxidation rxn
if the electrons are on the reactant side of the equation => reduction rxn
=> give a look at the equations in your post... you should be able to see the electrons listed and assign oxidation or reduction definitions. now :-)
Answer:
1. H2(g)⟶2H+(aq)+2e−: oxidation.
2. 12O2(g)+2H+(aq)+2e−⟶H2O(g): reduction.
3. Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e−: oxidation
4. 2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq): reduction
5. Fe(s)⟶Fe2+(aq)+2e−: oxidation
Explanation:
Hello,
1. At first, we realize that in this chemical reaction the hydrogen was initially with zero as the oxidation state and since it increases to +1, one says that it is an oxidation chemical reaction.
2. Secondly, since oxygen was with zero as the oxidation state and after the reaction it is lowered to -2, one says that it is a reduction chemical reaction.
3. Next, cadmium was a single metal so its oxidation state is 0 but after the reaction it increases to +2, therefore it is about an oxidation chemical reaction.
4. Then, nickel was initially with +3 as the oxidation state but after the chemical reaction is dwindles to +2, therefore it is about a reduction chemical reaction.
5. Finally, since iron was initially pure its oxidation state is zero but after the chemical reaction it increases to +2, therefore it is about an oxidation chemical reaction.
Best regards.
