Answer:
C,
Explanation:
3Mg + Fe2O3 ---> 2Fe + 3MgO
this is a single replacement reaction.
Answer: N (the Nitrogen)
Explanation:
Reduction refers to a decrease in oxidation number/state due to the gaining of electrons. As such the species that is being reduced will show a decrease in oxidation state.
Based on the redox rules,
Zn(s) has oxidation number of 0 [<em>rule 1: the oxidation number of an element in its free (uncombined) state is zero</em>]
Zn²⁺ has oxidation number of +2 [<em>rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion</em>]
Now, since Nitrogen is enbedded in a polyatomic ion in both cases, you have to do a bit a calculation to obtain the oxidation state.
For NO₃⁻ : N + (-2 × 3) = -1
N - 6 = -1
N = 5
<em>[Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 6: The oxidation state of hydrogen in a compound is usually +1]</em>
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For NH₄⁺ :
N + (4 x 1) = 1
N + 4 = 1
N = -3
[<em>Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 5: The oxidation number of oxygen in a compound is usually –2]</em>
Therefore, Zn moves from oxidation state of 0 to +2 (oxidation), while N moves from +5 to -3 (reduction).
Answer:
E°cell= 2.00 V
Explanation:
Let's consider the following reductions with their respective standard reduction potentials.
Al³⁺ + 3 e⁻ → Al; E°red= −1.66 V
Cu²⁺ + 2 e⁻ → Cu; E°red= 0.340 V
The one with the higher standard reduction potential will occur as a reduction (cathode) while the other will occur as an oxidation (anode). The corresponding half-reactions are:
Anode (oxidation): Al → Al³⁺ + 3 e⁻; E°red= −1.66 V
Cathode (reduction): Cu²⁺ + 2 e⁻ → Cu; E°red= 0.340 V
The standard cell potential (E°cell) can be calculated by subtracting the standard half-cell potential of the reaction that occurs at the cathode from the standard half-cell potential of the reaction that occurs at the anode:
E°cell=E°red(reduction process)−E°red(oxidation process)
= 0.340V - (-1.66V) = 2.00V
