Question
Which change involves oxidation of N?
A. \(NH_3\) to \(N_2\)
B. \(NO_2\) to NO
C. \(N_2\) to AlN
D. \(NO_2\) to \(N_2O_4\)
▶️Answer/Explanation
Markscheme: A
\(\mathrm{NH}_3\) to \(\mathrm{N}_2\)
1. Oxidation State of Nitrogen in \(\mathrm{NH}_3\) : In ammonia \(\left(\mathrm{NH}_3\right)\), nitrogen has an oxidation state of -3. Each hydrogen has an oxidation state of +1 , and there are three hydrogen atoms.
2. Oxidation State of Nitrogen in \(N_2\) : In dinitrogen \(\left(N_2\right)\), nitrogen has an oxidation state of 0 . Each nitrogen atom contributes zero oxidation state.
So, the change involves nitrogen going from an oxidation state of – 3 in \(\mathrm{NH}_3\) to \(\mathrm{O}\) in \(\mathrm{N}_2\). Nitrogen is gaining more positive oxidation state, and this is characteristic of oxidation. The
Question
Which combination describes an electrolytic cell?
▶️Answer/Explanation
Markscheme: B
An electrolytic cell is a type of electrochemical cell that uses an external source of electrical energy to drive a non-spontaneous reaction. In an electrolytic cell, electrical energy is used to cause a chemical change, typically by forcing a non-spontaneous redox reaction to occur.
Here’s a breakdown of the terms:
Electrical to Chemical: In an electrolytic cell, electrical energy is converted into chemical energy. The external electrical source provides the energy necessary to drive a non-spontaneous chemical reaction.
Spontaneity: A non-spontaneous process is one that does not occur on its own, and it requires an input of energy to proceed. In the context of an electrolytic cell, the non-spontaneous chemical reaction is made to happen by the input of electrical energy.
So, in summary, the electrolytic cell takes electrical energy from an external source and uses it to drive a non-spontaneous chemical reaction, converting electrical energy into chemical energy. The overall process is not spontaneous and requires an external energy source to occur.
Question
What is the standard potential difference of this cell?
\(Fe(s) | Fe^{2+}(aq) || Ag^+(aq) | Ag(s)\)
A. 0.45 + 2(0.80)
B. 0.45 + 0.80
C. -0.45 – 2(0.80)
D. -0.45 – 0.80
▶️Answer/Explanation
Markscheme: B
The standard cell potential (\(E^\ominus\)) for a cell is calculated by subtracting the standard reduction potential of the half-reaction at the anode from the standard reduction potential of the half-reaction at the cathode.
In this cell:
\[\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(\mathrm{aq})\|\mathrm{Ag}^{+}(\mathrm{aq})\right|\mathrm{Ag}(\mathrm{s})\]
The anode half-reaction is \(\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}(\mathrm{s})\) with a standard reduction potential of \(-0.45 \mathrm{~V}\).
The cathode half-reaction is \(\mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})\) with a standard reduction potential of \(+0.80 \mathrm{~V}\).
The overall standard cell potential (\(E^\ominus\)) is given by:
\[E^\ominus = E^\ominus_{\text{cathode}} – E^\ominus_{\text{anode}}\]
\[E^\ominus = (+0.80 \mathrm{~V}) – (-0.45 \mathrm{~V})\]
\[E^\ominus = 0.80 + 0.45\]
So, the correct answer is:
B. \(0.45 + 0.80\)
Question
Which statement is correct about the ions in a cell assembled from these half-cells?
A. Negative ions flow into the zinc half-cell from the salt bridge.
B. Negative ions flow into the nickel half-cell from the salt bridge.
C. \(Zn^{2+}\) ions are reduced to Zn.
D. The concentration of \(Ni^{2+}\) ions increases.
▶️Answer/Explanation
Markscheme: A
The correct statement can be determined by looking at the sign of the standard cell potential (\(E^\ominus\)) for the overall cell reaction.
The overall cell reaction is obtained by adding the two half-reactions:
\[\mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{e}^- \rightarrow \mathrm{Ni}(\mathrm{s}) \quad \text{(cathode)}\]
\[\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + 2\mathrm{e}^- \quad \text{(anode)}\]
The standard cell potential (\(E^\ominus\)) is given by the cathode potential minus the anode potential:
\[E^\ominus = E^\ominus_{\text{cathode}} – E^\ominus_{\text{anode}}\]
\[E^\ominus = (-0.26 \mathrm{~V}) – (-0.76 \mathrm{~V})\]
\[E^\ominus = 0.50 \mathrm{~V}\]
Since the standard cell potential is positive (0.50 V), the reaction is spontaneous as written. In a spontaneous cell reaction, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs).
Now, looking at the options:
A. Negative ions flow into the zinc half-cell from the salt bridge.
This is correct. In a spontaneous cell, electrons flow from the anode to the cathode. In the zinc half-cell (the anode), zinc undergoes oxidation (\(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^-\)), so negative ions (\(\mathrm{Zn}^{2+}\)) move into the zinc half-cell from the salt bridge to balance the charge.
Therefore, the correct answer is A.
Question
Which species is the oxidizing agent?
\(14H^+(aq) + 2Mn^{2+}(aq) \rightarrow 2MnO_4^-(aq) + 5Bi^{3+}(aq) + 7H_2O(l)\)
A. \(H^+(aq)\)
B. \(Mn^{2+}(aq)\)
C. \(BiO_3^-(aq)\)
D. \(MnO_4^-(aq)\)
▶️Answer/Explanation
Markscheme: C
To determine the oxidizing agent, we need to identify the species that undergoes reduction (loses electrons) in the given reaction. The oxidizing agent is the species that causes another substance to be oxidized.
In the reaction:
\[14H^+(aq) + 2Mn^{2+}(aq) \rightarrow 2MnO_4^-(aq) + 5Bi^{3+}(aq) + 7H_2O(l)\]
1. Oxidation state changes:
The oxidation state of manganese (\(Mn^{2+}\)) increases from +2 to +7.
The oxidation state of bismuth (\(Bi^{3+}\)) remains the same.
2. Oxidation agent: The species that causes another substance to be oxidized is the one undergoing reduction. In this case, \(Bi^{3+}\) is being reduced (its oxidation state remains the same), so the oxidizing agent is \(BiO_3^-(aq)\).
Therefore, the correct answer is C. \(BiO_3^-(aq)\).