101. The standard reduction potentials are:
\[E^\circ\left(Mg^{2+}/Mg\right) = -2.37\,\text{V}\]
\[E^\circ\left(Cu^{2+}/Cu\right) = +0.34\,\text{V}\]
Which overall reaction is spontaneous under standard conditions?
ⓐ. \(Cu + Mg^{2+} \rightarrow Cu^{2+} + Mg\)
ⓑ. \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\)
ⓒ. \(Mg + Cu \rightarrow Mg^{2+} + Cu^{2+}\)
ⓓ. \(Mg^{2+} + Cu^{2+} \rightarrow Mg + Cu\)
Correct Answer: \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\)
Explanation: The half-cell with the higher standard reduction potential undergoes reduction, so \(Cu^{2+}\) is reduced to \(Cu\). Magnesium has the lower standard reduction potential, so it undergoes oxidation to \(Mg^{2+}\). Combining these two favorable half-reactions gives \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\) as the spontaneous reaction.
102. The standard reduction potentials are:
\[E^\circ\left(Pb^{2+}/Pb\right) = -0.13\,\text{V}\]
\[E^\circ\left(Fe^{2+}/Fe\right) = -0.44\,\text{V}\]
Which statement is correct for the galvanic cell formed by these half-cells?
ⓐ. \(Pb^{2+}/Pb\) is the cathode and electrons flow from \(Fe\) to \(Pb\)
ⓑ. \(Fe^{2+}/Fe\) is the cathode and electrons flow from \(Pb\) to \(Fe\)
ⓒ. \(Pb^{2+}/Pb\) is the anode and electrons flow from \(Fe\) to \(Pb\)
ⓓ. Both half-cells act as cathodes because both potentials are negative
Correct Answer: \(Pb^{2+}/Pb\) is the cathode and electrons flow from \(Fe\) to \(Pb\)
Explanation: The cathode is the half-cell with the higher standard reduction potential. Since \(-0.13\,\text{V}\) is higher than \(-0.44\,\text{V}\), the \(Pb^{2+}/Pb\) half-cell acts as the cathode. The iron half-cell is the anode, so electrons flow from \(Fe\) to \(Pb\) through the external circuit.
103. If two half-cells have exactly the same standard reduction potential, what is the value of \(E^\circ_{\text{cell}}\) for the cell formed from them under standard conditions?
ⓐ. It must be positive and very small
ⓑ. It must be negative and very small
ⓒ. It depends only on the salt bridge used
ⓓ. \(0\text{ V}\)
Correct Answer: \(0\text{ V}\)
Explanation: Standard cell emf is the difference between the standard reduction potentials of the cathode and anode. If both half-cells have the same value, their difference is zero. So no net driving force exists under standard conditions, and \(E^\circ_{\text{cell}} = 0.00\,\text{V}\).
104. An unknown half-cell is connected to the standard hydrogen electrode under standard conditions. The standard hydrogen electrode acts as the cathode, and the cell emf is \(0.28\,\text{V}\). What is the standard reduction potential of the unknown half-cell?
ⓐ. \(+0.28\,\text{V}\)
ⓑ. \(0.00\,\text{V}\)
ⓒ. \(-0.28\,\text{V}\)
ⓓ. \(+0.56\,\text{V}\)
Correct Answer: \(-0.28\,\text{V}\)
Explanation: \(\textbf{Given:}\)
\[E^\circ_{\text{cell}} = 0.28\,\text{V}\]
For the standard hydrogen electrode, \(E^\circ = 0.00\,\text{V}\)
\(\textbf{Required:}\)
Standard reduction potential of the unknown half-cell
\(\textbf{Relevant formula:}\)
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]
\(\textbf{Why this formula applies:}\)
The standard hydrogen electrode is acting as the cathode, so the unknown half-cell is the anode.
\(\textbf{Identify known values:}\)
\[E^\circ_{\text{cathode}} = 0.00\,\text{V}\]
\[E^\circ_{\text{anode}} = ?\]
\(\textbf{Substitution:}\)
\[0.28 = 0.00 - E^\circ_{\text{anode}}\]
\(\textbf{Rearrangement:}\)
\[E^\circ_{\text{anode}} = -0.28\,\text{V}\]
\(\textbf{Interpretation:}\)
Since the unknown half-cell is the anode, this value is also its standard reduction potential.
\(\textbf{Unit check:}\)
Electrode potential is expressed in volt.
\(\textbf{Final Answer:}\)
The standard reduction potential of the unknown half-cell is \(-0.28\,\text{V}\).
105. A student uses standard reduction potentials and writes \(E^\circ_{\text{cell}} = E^\circ_{\text{anode}} - E^\circ_{\text{cathode}}\). Which correction is required?
ⓐ. The formula should be \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
ⓑ. The formula should be \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} + E^\circ_{\text{anode}}\)
ⓒ. The formula should be \(E^\circ_{\text{cell}} = \dfrac{E^\circ_{\text{cathode}}}{E^\circ_{\text{anode}}}\)
ⓓ. No correction is required if both values are reduction potentials
Correct Answer: The formula should be \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
Explanation: Standard electrode potentials are tabulated as reduction potentials. For a galvanic cell, the cathode is the half-cell where reduction occurs and the anode is the half-cell where oxidation occurs. Therefore the correct relation is obtained by subtracting the reduction potential of the anode from that of the cathode:
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\].
106. The standard reduction potentials are:
\[E^\circ\left(Sn^{2+}/Sn\right) = -0.14\,\text{V}\]
\[E^\circ\left(Pb^{2+}/Pb\right) = -0.13\,\text{V}\]
What is the standard emf of the galvanic cell formed by these half-cells?
ⓐ. \(-0.01\,\text{V}\)
ⓑ. \(+0.27\,\text{V}\)
ⓒ. \(+0.01\,\text{V}\)
ⓓ. \(-0.27\,\text{V}\)
Correct Answer: \(+0.01\,\text{V}\)
Explanation: \(\textbf{Given:}\)
\[E^\circ\left(Sn^{2+}/Sn\right) = -0.14\,\text{V}\]
\[E^\circ\left(Pb^{2+}/Pb\right) = -0.13\,\text{V}\]
\(\textbf{Required:}\)
\[E^\circ_{\text{cell}}\]
\(\textbf{Relevant formula:}\)
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]
\(\textbf{Why this formula applies:}\)
The half-cell with the higher reduction potential acts as the cathode.
\(\textbf{Identify cathode and anode:}\)
Since \(-0.13\,\text{V}\) is higher than \(-0.14\,\text{V}\),
Cathode: \(Pb^{2+}/Pb\)
Anode: \(Sn^{2+}/Sn\)
\(\textbf{Substitution:}\)
\[E^\circ_{\text{cell}} = (-0.13) - (-0.14)\]
\(\textbf{Intermediate simplification:}\)
\[E^\circ_{\text{cell}} = -0.13 + 0.14\]
\(\textbf{Final simplification:}\)
\[E^\circ_{\text{cell}} = +0.01\,\text{V}\]
\(\textbf{Unit check:}\)
Emf is expressed in volt.
\(\textbf{Final Answer:}\)
\[E^\circ_{\text{cell}} = +0.01\,\text{V}\].
107. The standard reduction potentials are:
\[E^\circ\left(Al^{3+}/Al\right) = -1.66\,\text{V}\]
\[E^\circ\left(Fe^{2+}/Fe\right) = -0.44\,\text{V}\]
Which reaction is spontaneous under standard conditions?
ⓐ. \(Fe + Al^{3+} \rightarrow Fe^{2+} + Al\)
ⓑ. \(2Al + 3Fe^{2+} \rightarrow 2Al^{3+} + 3Fe\)
ⓒ. \(Al + Fe \rightarrow Al^{3+} + Fe^{2+}\)
ⓓ. \(2Al^{3+} + 3Fe^{2+} \rightarrow 2Al + 3Fe\)
Correct Answer: \(2Al + 3Fe^{2+} \rightarrow 2Al^{3+} + 3Fe\)
Explanation: The half-cell with higher standard reduction potential undergoes reduction more readily. Since \(-0.44\,\text{V}\) is higher than \(-1.66\,\text{V}\), \(Fe^{2+}\) is reduced to \(Fe\). Aluminium therefore undergoes oxidation to \(Al^{3+}\). Combining the balanced half-reactions gives:
\[2Al + 3Fe^{2+} \rightarrow 2Al^{3+} + 3Fe\].
108. If a cell reaction has \(E^\circ_{\text{cell}} = +0.52\,\text{V}\) as written, then for the reverse reaction the standard emf will be
ⓐ. \(+1.04\,\text{V}\)
ⓑ. \(0.00\,\text{V}\)
ⓒ. \(+0.52\,\text{V}\)
ⓓ. \(-0.52\,\text{V}\)
Correct Answer: \(-0.52\,\text{V}\)
Explanation: Reversing the cell reaction reverses the direction of spontaneity. The magnitude of the standard emf remains the same, but its sign changes. Therefore if the forward reaction has \(E^\circ_{\text{cell}} = +0.52\,\text{V}\), the reverse reaction has \(E^\circ_{\text{cell}} = -0.52\,\text{V}\).
109. The standard reduction potentials are:
\[E^\circ\left(X^{2+}/X\right) = +0.20\,\text{V}\]
\[E^\circ\left(Y^{2+}/Y\right) = -0.40\,\text{V}\]
Which statement is correct for the galvanic cell formed by these half-cells?
ⓐ. \(X\) is oxidized and electrons flow from \(X\) to \(Y\)
ⓑ. \(Y\) is reduced and electrons flow from \(X\) to \(Y\)
ⓒ. \(X^{2+}\) is reduced and electrons flow from \(Y\) to \(X\)
ⓓ. \(Y^{2+}\) is reduced and electrons flow from \(Y\) to \(X\)
Correct Answer: \(X^{2+}\) is reduced and electrons flow from \(Y\) to \(X\)
Explanation: The higher standard reduction potential corresponds to the half-cell that is reduced more readily. Since \(+0.20\,\text{V}\) is higher than \(-0.40\,\text{V}\), the \(X^{2+}/X\) half-cell is the cathode, so \(X^{2+}\) is reduced. The \(Y^{2+}/Y\) half-cell is the anode, so electrons flow from \(Y\) to \(X\) through the external circuit.
110. A galvanic cell is formed by two half-cells having standard reduction potentials \(-0.80\,\text{V}\) and \(-0.25\,\text{V}\). Which statement is correct?
ⓐ. The half-cell with \(-0.80\,\text{V}\) is the cathode because it is more negative.
ⓑ. The half-cell with \(-0.25\,\text{V}\) is the cathode.
ⓒ. Both half-cells act as cathodes because both values are negative.
ⓓ. Cathode and anode cannot be predicted when both values are negative.
Correct Answer: The half-cell with \(-0.25\,\text{V}\) is the cathode.
Explanation: The sign alone is not enough; comparison is what matters. Among two standard reduction potentials, the larger value corresponds to the greater reduction tendency. Since \(-0.25\,\text{V}\) is greater than \(-0.80\,\text{V}\), the half-cell with \(-0.25\,\text{V}\) acts as the cathode.
111. The standard reduction potentials are:
\[E^\circ\left(M^{2+}/M\right) = +0.50\,\text{V}\]
\[E^\circ\left(N^{2+}/N\right) = +0.10\,\text{V}\]
What is the standard emf of the galvanic cell formed by these half-cells?
ⓐ. \(+0.60\,\text{V}\)
ⓑ. \(-0.60\,\text{V}\)
ⓒ. \(-0.40\,\text{V}\)
ⓓ. \(+0.40\,\text{V}\)
Correct Answer: \(+0.40\,\text{V}\)
Explanation: \(\textbf{Given:}\)
\[E^\circ\left(M^{2+}/M\right) = +0.50\,\text{V}\]
\[E^\circ\left(N^{2+}/N\right) = +0.10\,\text{V}\]
\(\textbf{Required:}\)
\[E^\circ_{\text{cell}}\]
\(\textbf{Relevant formula:}\)
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]
\(\textbf{Why this formula applies:}\)
The half-cell with the greater reduction potential acts as the cathode.
\(\textbf{Identify cathode and anode:}\)
Cathode: \(M^{2+}/M\)
Anode: \(N^{2+}/N\)
\(\textbf{Substitution:}\)
\[E^\circ_{\text{cell}} = 0.50 - 0.10\]
\(\textbf{Final simplification:}\)
\[E^\circ_{\text{cell}} = 0.40\,\text{V}\]
\(\textbf{Unit check:}\)
The unit of emf is volt.
\(\textbf{Final Answer:}\)
\[E^\circ_{\text{cell}} = +0.40\,\text{V}\].
112. Which statement best links the sign of \(E^\circ_{\text{cell}}\) with the feasibility of the reaction under standard conditions?
ⓐ. A negative value means the reaction is spontaneous as written.
ⓑ. A positive value means the forward reaction is feasible.
ⓒ. A zero value means the reaction will proceed rapidly in both directions.
ⓓ. A positive value means the reverse reaction is the only possible one.
Correct Answer: A positive value means the forward reaction is feasible.
Explanation: Standard cell emf indicates whether the reaction as written can proceed spontaneously under standard conditions. A positive value supports spontaneity in the forward direction, while a negative value means the reverse direction is favored. A zero value corresponds to equilibrium tendency.
113. The standard reduction potentials are:
\[E^\circ\left(Ag^+/Ag\right) = +0.80\,\text{V}\]
\[E^\circ\left(Cu^{2+}/Cu\right) = +0.34\,\text{V}\]
\[E^\circ\left(Zn^{2+}/Zn\right) = -0.76\,\text{V}\]
\[E^\circ\left(Mg^{2+}/Mg\right) = -2.37\,\text{V}\]
Which species is the strongest oxidizing agent?
ⓐ. \(Ag^+\)
ⓑ. \(Cu^{2+}\)
ⓒ. \(Zn^{2+}\)
ⓓ. \(Mg^{2+}\)
Correct Answer: \(Ag^+\)
Explanation: An oxidizing agent gets reduced. The stronger oxidizing agent has the greater tendency to gain electrons, so it has the highest standard reduction potential. Among the given species, \(Ag^+\) has the highest value of \(+0.80\,\text{V}\), so it is the strongest oxidizing agent.
114. Using the same data, which metal is the strongest reducing agent?
ⓐ. \(Ag\)
ⓑ. \(Cu\)
ⓒ. \(Mg\)
ⓓ. \(Zn\)
Correct Answer: \(Mg\)
Explanation: A reducing agent loses electrons and gets oxidized. The metal whose corresponding reduction potential is most negative has the greatest tendency to undergo oxidation. Since \(E^\circ\left(Mg^{2+}/Mg\right) = -2.37\,\text{V}\) is the most negative, \(Mg\) is the strongest reducing agent.
115. Which metal can displace \(Ag^+\) from its solution but cannot displace \(Zn^{2+}\) from its solution?
ⓐ. \(Ag\)
ⓑ. \(Cu\)
ⓒ. \(Mg\)
ⓓ. \(Pt\)
Correct Answer: \(Cu\)
Explanation: A metal can displace the ions of a metal that has a higher standard reduction potential. Copper can reduce \(Ag^+\) because \(E^\circ\left(Ag^+/Ag\right)\) is higher than \(E^\circ\left(Cu^{2+}/Cu\right)\). But copper cannot displace \(Zn^{2+}\) because zinc has a lower standard reduction potential and is more easily oxidized than copper.
116. Which order correctly represents increasing oxidizing strength of the ions?
ⓐ. \(Ag^+ < Cu^{2+} < Zn^{2+} < Mg^{2+}\)
ⓑ. \(Zn^{2+} < Mg^{2+} < Cu^{2+} < Ag^+\)
ⓒ. \(Cu^{2+} < Ag^+ < Zn^{2+} < Mg^{2+}\)
ⓓ. \(Mg^{2+} < Zn^{2+} < Cu^{2+} < Ag^+\)
Correct Answer: \(Mg^{2+} < Zn^{2+} < Cu^{2+} < Ag^+\)
Explanation: Oxidizing strength increases with increasing standard reduction potential because a better oxidizing agent is reduced more readily. Here the reduction potentials increase from \(Mg^{2+}/Mg\) to \(Ag^+/Ag\). Therefore the oxidizing strength follows
\[Mg^{2+} < Zn^{2+} < Cu^{2+} < Ag^+\].
117. Which order correctly represents increasing reducing strength of the metals?
ⓐ. \(Mg < Zn < Cu < Ag\)
ⓑ. \(Ag < Cu < Zn < Mg\)
ⓒ. \(Cu < Ag < Mg < Zn\)
ⓓ. \(Zn < Mg < Ag < Cu\)
Correct Answer: \(Ag < Cu < Zn < Mg\)
Explanation: Reducing strength of a metal depends on how easily it loses electrons. A metal becomes a stronger reducing agent as the standard reduction potential of its ion becomes more negative. Thus the reducing strength increases in the order
\[Ag < Cu < Zn < Mg\].
118. Which statement about the electrochemical series is correct?
ⓐ. A species with more positive standard reduction potential is more easily reduced.
ⓑ. A species with more negative standard reduction potential is always the stronger oxidizing agent.
ⓒ. A metal ion with lower standard reduction potential is always reduced first.
ⓓ. A metal with higher standard reduction potential is always the stronger reducing agent.
Correct Answer: A species with more positive standard reduction potential is more easily reduced.
Explanation: Standard reduction potential measures the tendency of a species to gain electrons. A more positive value means reduction is more favorable. That is why such a species acts more readily as an oxidizing agent.
119. A metal \(X\) is dipped into a solution containing \(Y^{2+}\) ions, and a deposit of \(Y\) forms on the metal surface. What is the best conclusion?
ⓐ. \(X\) has a higher reduction tendency than \(Y^{2+}\).
ⓑ. \(Y\) is a stronger reducing agent than \(X\).
ⓒ. \(X\) acts as the cathode and gets reduced.
ⓓ. \(X\) loses electrons more readily than \(Y\).
Correct Answer: \(X\) loses electrons more readily than \(Y\).
Explanation: If \(Y\) is deposited, then \(Y^{2+}\) is reduced to \(Y\). The electrons needed for that reduction must come from \(X\), so \(X\) is oxidized. Therefore \(X\) has the greater tendency to lose electrons and acts as the stronger reducing agent.
120. Metal \(M\) displaces copper from \(CuSO_4\) solution, but copper does not displace \(M\) from a solution containing \(M^{2+}\). Which statement is correct?
ⓐ. \(M\) and \(Cu\) have the same standard reduction potential.
ⓑ. \(M\) is less reactive than copper.
ⓒ. \(M^{2+}/M\) has the lower \(E^\circ\).
ⓓ. \(M^{2+}\) is a stronger oxidizing agent than \(Cu^{2+}\).
Correct Answer: \(M^{2+}/M\) has the lower \(E^\circ\).
Explanation: If \(M\) displaces copper from \(CuSO_4\), then \(M\) is oxidized while \(Cu^{2+}\) is reduced. This means \(M\) has the greater tendency to lose electrons. Therefore the standard reduction potential of \(M^{2+}/M\) must be lower than that of \(Cu^{2+}/Cu\).
