401. A cell is made from \( \mathrm{M^{2+}/M} \) and the standard hydrogen electrode. If \(E^\circ_{\mathrm{M^{2+}/M}}=-0.40\,\mathrm{V}\), the reaction \( \mathrm{M+2H^+\rightarrow M^{2+}+H_2} \) has:
ⓐ. \(E^\circ_{\text{cell}}=+0.40\,\mathrm{V}\)
ⓑ. \(E^\circ_{\text{cell}}=-0.40\,\mathrm{V}\)
ⓒ. \(E^\circ_{\text{cell}}=+0.80\,\mathrm{V}\)
ⓓ. \(E^\circ_{\text{cell}}=0.00\,\mathrm{V}\)
Correct Answer: \(E^\circ_{\text{cell}}=+0.40\,\mathrm{V}\)
Explanation: \( \textbf{Reaction direction:} \) \( \mathrm{M} \) is oxidised to \( \mathrm{M^{2+}} \), so the metal half-cell is the anode.
\( \textbf{Hydrogen half-cell:} \) \( \mathrm{2H^++2e^-\rightarrow H_2} \) is the cathode reduction.
\( \textbf{Standard potentials:} \)
\[
E^\circ_{\mathrm{H^+/H_2}}=0.00\,\mathrm{V}
\]
\[
E^\circ_{\mathrm{M^{2+}/M}}=-0.40\,\mathrm{V}
\]
\( \textbf{Cell potential:} \)
\[
E^\circ_{\text{cell}}=0.00-(-0.40)
\]
\[
E^\circ_{\text{cell}}=+0.40\,\mathrm{V}
\]
\( \textbf{Final answer:} \) The reaction has \(E^\circ_{\text{cell}}=+0.40\,\mathrm{V}\). A negative metal reduction potential can still give a positive cell potential when paired with hydrogen ion reduction.
402. A metal \( \mathrm{N} \) does not liberate \( \mathrm{H_2} \) from dilute acid under standard comparison, and \(E^\circ_{\mathrm{N^{2+}/N}}=+0.20\,\mathrm{V}\). The best explanation is:
ⓐ. \( \mathrm{N} \) has no electrons
ⓑ. hydrogen electrode has no redox half-reaction
ⓒ. positive reduction potential always means a metal dissolves in acid
ⓓ. \( \mathrm{N^{2+}} \) is easier to reduce than \( \mathrm{H^+} \)
Correct Answer: \( \mathrm{N^{2+}} \) is easier to reduce than \( \mathrm{H^+} \)
Explanation: The standard hydrogen electrode has \(E^\circ=0.00\,\mathrm{V}\). If \(E^\circ_{\mathrm{N^{2+}/N}}\) is \(+0.20\,\mathrm{V}\), the reduction of \( \mathrm{N^{2+}} \) is more favourable than reduction of \( \mathrm{H^+} \). For \( \mathrm{N} \) to liberate hydrogen, \( \mathrm{N} \) would need to act as the reducing agent and \( \mathrm{H^+} \) would need to be reduced. The cell potential for that written acid reaction would be \(0.00-(+0.20)=-0.20\,\mathrm{V}\). The positive reduction potential of the metal ion therefore does not support spontaneous hydrogen liberation.
403. A table lists half-cell data.
| Half-cell | \(E^\circ\) |
|---|
| \( \mathrm{A^{2+}+2e^-\rightarrow A} \) | \(-1.00\,\mathrm{V}\) |
| \( \mathrm{B^{2+}+2e^-\rightarrow B} \) | \(-0.20\,\mathrm{V}\) |
| \( \mathrm{C^{2+}+2e^-\rightarrow C} \) | \(+0.60\,\mathrm{V}\) |
The strongest reducing metal and strongest oxidising ion are respectively:
ⓐ. \( \mathrm{C} \) and \( \mathrm{A^{2+}} \)
ⓑ. \( \mathrm{A} \) and \( \mathrm{C^{2+}} \)
ⓒ. \( \mathrm{B} \) and \( \mathrm{B^{2+}} \)
ⓓ. \( \mathrm{A^{2+}} \) and \( \mathrm{C} \)
Correct Answer: \( \mathrm{A} \) and \( \mathrm{C^{2+}} \)
Explanation: The strongest reducing metal is the one most easily oxidised. This corresponds to the metal whose ion has the most negative reduction potential, because the reverse oxidation is most favoured. \( \mathrm{A^{2+}/A} \) has the most negative value, so \( \mathrm{A} \) is the strongest reducing metal. The strongest oxidising ion is the ion most easily reduced. \( \mathrm{C^{2+}} \) has the most positive reduction potential, so it is the strongest oxidising ion.
404. According to the data in the table, the standard reaction expected to have the largest positive cell potential is:
ⓐ. \( \mathrm{C+A^{2+}\rightarrow C^{2+}+A} \)
ⓑ. \( \mathrm{A+C^{2+}\rightarrow A^{2+}+C} \)
ⓒ. \( \mathrm{B+A^{2+}\rightarrow B^{2+}+A} \)
ⓓ. \( \mathrm{C+B^{2+}\rightarrow C^{2+}+B} \)
Correct Answer: \( \mathrm{A+C^{2+}\rightarrow A^{2+}+C} \)
Explanation: \( \textbf{Best anode choice:} \) The strongest reducing metal is \( \mathrm{A} \), whose reduction potential is \(-1.00\,\mathrm{V}\).
\( \textbf{Best cathode choice:} \) The strongest oxidising ion is \( \mathrm{C^{2+}} \), whose reduction potential is \(+0.60\,\mathrm{V}\).
\( \textbf{Cell potential:} \)
\[
E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}
\]
\[
E^\circ_{\text{cell}}=(+0.60)-(-1.00)=+1.60\,\mathrm{V}
\]
\( \textbf{Reaction:} \)
\[
\mathrm{A+C^{2+}\rightarrow A^{2+}+C}
\]
\( \textbf{Final answer:} \) The largest positive value comes from pairing the easiest oxidation with the easiest reduction. Using two closer potentials would give a smaller driving tendency.
405. A proposed galvanic cell has \(E^\circ_{\text{cell}}=0.00\,\mathrm{V}\) under standard conditions. The safest interpretation is:
ⓐ. no net standard driving force in the written direction
ⓑ. the equation must contain no atoms
ⓒ. oxidation and reduction cannot be written as half-reactions
ⓓ. the cell potential must be multiplied by coefficients to become positive
Correct Answer: no net standard driving force in the written direction
Explanation: A positive \(E^\circ_{\text{cell}}\) indicates a spontaneous written reaction under standard conditions. A negative value indicates the reverse direction is favoured. A value of \(0.00\,\mathrm{V}\) means there is no net standard driving force for the written direction compared with its reverse. It does not mean atoms or half-reactions are absent. Coefficients do not turn a zero potential into a positive potential because electrode potential is not scaled by reaction multiplication.
406. A reaction has \(E^\circ_{\text{cell}}=+0.50\,\mathrm{V}\). If the reaction is reversed, the standard cell potential becomes:
ⓐ. \(+0.50\,\mathrm{V}\)
ⓑ. \(+1.00\,\mathrm{V}\)
ⓒ. \(0.00\,\mathrm{V}\)
ⓓ. \(-0.50\,\mathrm{V}\)
Correct Answer: \(-0.50\,\mathrm{V}\)
Explanation: Reversing a redox reaction swaps the oxidation and reduction directions. The cathode of the original direction becomes the anode in the reverse direction, and the anode becomes the cathode. Therefore the sign of \(E^\circ_{\text{cell}}\) changes. The magnitude remains \(0.50\,\mathrm{V}\) if only the direction is reversed. This sign reversal is different from multiplying all coefficients, which does not change the potential.
407. A graph plots standard reduction potential on the y-axis for \( \mathrm{A^{2+}/A} \), \( \mathrm{B^{2+}/B} \), and \( \mathrm{C^{2+}/C} \). The points are at \(-0.90\,\mathrm{V}\), \(+0.10\,\mathrm{V}\), and \(+0.80\,\mathrm{V}\), respectively. The pair giving the greatest \(E^\circ_{\text{cell}}\) is:
ⓐ. \( \mathrm{C/C^{2+}} \) as anode and \( \mathrm{A^{2+}/A} \) as cathode
ⓑ. \( \mathrm{A/A^{2+}} \) as anode and \( \mathrm{C^{2+}/C} \) as cathode
ⓒ. \( \mathrm{B/B^{2+}} \) as anode and \( \mathrm{A^{2+}/A} \) as cathode
ⓓ. \( \mathrm{C/C^{2+}} \) as anode and \( \mathrm{B^{2+}/B} \) as cathode
Correct Answer: \( \mathrm{A/A^{2+}} \) as anode and \( \mathrm{C^{2+}/C} \) as cathode
Explanation: \( \textbf{Greatest cathode value:} \) The highest reduction potential is \(+0.80\,\mathrm{V}\) for \( \mathrm{C^{2+}/C} \).
\( \textbf{Lowest anode reduction value:} \) The lowest reduction potential is \(-0.90\,\mathrm{V}\) for \( \mathrm{A^{2+}/A} \).
\( \textbf{Cell potential:} \)
\[
E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}
\]
\[
E^\circ_{\text{cell}}=(+0.80)-(-0.90)=+1.70\,\mathrm{V}
\]
\( \textbf{Final answer:} \) The greatest cell potential comes from using \( \mathrm{A} \) as the oxidised metal at the anode and \( \mathrm{C^{2+}} \) as the reduced ion at the cathode. The widest separation between reduction potentials gives the largest positive value.
408. A student sees \(E^\circ_{\mathrm{Al^{3+}/Al}}=-1.66\,\mathrm{V}\) and says aluminium ion is the strongest oxidising agent because the magnitude is large. The best correction is:
ⓐ. \( \mathrm{Al^{3+}} \) is hard to reduce; \( \mathrm{Al} \) is reductant
ⓑ. negative \(E^\circ\) makes \( \mathrm{Al^{3+}} \) the strongest oxidising agent
ⓒ. magnitude alone decides oxidising strength without sign
ⓓ. aluminium cannot take part in redox reactions
Correct Answer: \( \mathrm{Al^{3+}} \) is hard to reduce; \( \mathrm{Al} \) is reductant
Explanation: Standard reduction potential must be read with its sign. A more positive value means stronger tendency for reduction and stronger oxidising behaviour of the ion. A very negative value means the ion is relatively difficult to reduce under standard comparison. The reverse metal oxidation is comparatively favourable, so aluminium metal acts as a strong reducing agent. Looking only at numerical magnitude would reverse the chemical meaning of the table.
409. When a standard reduction-potential table is used to predict displacement reactions, a metal \( \mathrm{M} \) can reduce \( \mathrm{N^{2+}} \) if:
ⓐ. \(E^\circ_{\mathrm{M^{2+}/M}}\) is greater than \(E^\circ_{\mathrm{N^{2+}/N}}\)
ⓑ. both reduction potentials are exactly equal
ⓒ. both ions have the same charge only
ⓓ. \(E^\circ_{\mathrm{N^{2+}/N}}\) is greater than \(E^\circ_{\mathrm{M^{2+}/M}}\)
Correct Answer: \(E^\circ_{\mathrm{N^{2+}/N}}\) is greater than \(E^\circ_{\mathrm{M^{2+}/M}}\)
Explanation: If \( \mathrm{M} \) reduces \( \mathrm{N^{2+}} \), then \( \mathrm{M} \) is oxidised and \( \mathrm{N^{2+}} \) is reduced. For this to be favoured as a galvanic process, the cathode reduction potential must be greater than the anode reduction potential. The cathode is \( \mathrm{N^{2+}/N} \), and the anode couple is \( \mathrm{M^{2+}/M} \) as tabulated reduction. Thus \(E^\circ_{\mathrm{N^{2+}/N}} \gt E^\circ_{\mathrm{M^{2+}/M}}\). Equal ion charge alone cannot decide displacement.
410. Given \(E^\circ_{\mathrm{Pb^{2+}/Pb}}=-0.13\,\mathrm{V}\) and \(E^\circ_{\mathrm{Cu^{2+}/Cu}}=+0.34\,\mathrm{V}\), what happens when lead metal is placed in \( \mathrm{Cu^{2+}} \) solution under standard comparison?
ⓐ. \( \mathrm{Cu^{2+}} \) reduces \( \mathrm{Pb} \) to \( \mathrm{Pb^{2+}} \)
ⓑ. no redox direction can be compared from these values
ⓒ. \( \mathrm{Pb} \) reduces \( \mathrm{Cu^{2+}} \) to \( \mathrm{Cu} \)
ⓓ. \( \mathrm{Pb} \) must be reduced to \( \mathrm{Pb^{2+}} \)
Correct Answer: \( \mathrm{Pb} \) reduces \( \mathrm{Cu^{2+}} \) to \( \mathrm{Cu} \)
Explanation: Copper ion has the higher reduction potential, \(+0.34\,\mathrm{V}\), so it has greater tendency to be reduced than lead ion. Lead metal can act as the anode material and be oxidised to \( \mathrm{Pb^{2+}} \). The cell potential for \( \mathrm{Pb+Cu^{2+}\rightarrow Pb^{2+}+Cu} \) is \( (+0.34)-(-0.13)=+0.47\,\mathrm{V} \). A positive value supports the reaction as written under standard comparison. The wording “lead is reduced to \( \mathrm{Pb^{2+}} \)” is wrong because forming \( \mathrm{Pb^{2+}} \) is oxidation.
411. A voltmeter connected to a working \( \mathrm{Zn-Cu} \) galvanic cell shows a positive reading when the positive terminal is connected to the copper electrode. This is consistent with:
ⓐ. zinc electrode being the cathode
ⓑ. sulphate ions carrying electrons through the wire
ⓒ. copper electrode being the cathode
ⓓ. both electrodes undergoing oxidation
Correct Answer: copper electrode being the cathode
Explanation: In a \( \mathrm{Zn-Cu} \) galvanic cell, \( \mathrm{Cu^{2+}} \) is reduced at the copper electrode. The cathode is positive in a galvanic cell because electrons arrive there through the external circuit and are consumed by reduction. Zinc is oxidised at the anode and supplies electrons. A positive voltmeter terminal connected to the copper side therefore matches the cathode assignment. The wire carries electrons, while ions move through the solutions and salt bridge.
412. In an electrolytic cell, the anode is positive, but oxidation still occurs there. This shows that:
ⓐ. oxidation changes to reduction in electrolysis
ⓑ. cathode means oxidation in all cells
ⓒ. electrons are not involved in electrolysis
ⓓ. sign and process are separate
Correct Answer: sign and process are separate
Explanation: The anode is always the site of oxidation. In a galvanic cell the anode is negative, while in an electrolytic cell the anode is positive because an external power source drives the reaction. The electrode sign can therefore change with the type of cell. The process definition does not change: oxidation occurs at the anode and reduction occurs at the cathode. Separating sign from process prevents confusion when moving between galvanic and electrolytic examples.
413. The phrase “standard conditions” in standard electrode potential most directly means that:
ⓐ. agreed reference conditions for comparison
ⓑ. the reaction must always occur at the same speed
ⓒ. every solution must be colourless
ⓓ. all electrodes must have the same mass
Correct Answer: agreed reference conditions for comparison
Explanation: Standard electrode potentials are comparison values measured under agreed reference conditions. This allows different half-cells to be placed on the same scale. The purpose is to compare reduction tendencies in a consistent way. Standard conditions do not mean every reaction has the same rate or every electrode has the same mass. The value is thermodynamic in nature, not a direct measurement of reaction speed or colour.
414. A slow reaction has a positive \(E^\circ_{\text{cell}}\). The correct interpretation is:
ⓐ. it must be impossible
ⓑ. it cannot be a redox reaction
ⓒ. thermodynamically favoured but kinetically slow
ⓓ. positive voltage always means instantaneous reaction
Correct Answer: thermodynamically favoured but kinetically slow
Explanation: A positive standard cell potential indicates that the reaction is favoured under standard thermodynamic comparison. It does not guarantee that the reaction will be fast. Some reactions have favourable driving force but need suitable surface, catalyst, activation conditions, or time to proceed noticeably. Redox tendency and reaction rate are related to different aspects of a reaction. A voltage sign helps decide direction, not the exact speed of visible change.
415. Study the statement set.
I. More positive \(E^\circ\) for a reduction half-reaction means greater reduction tendency.
II. In a galvanic cell, the cathode has the reduction half-reaction.
III. When a half-reaction is doubled, its \(E^\circ\) value is doubled.
ⓐ. II and III only
ⓑ. I and II only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Statement I is true because standard reduction potential compares tendency to gain electrons. Statement II is true because reduction occurs at the cathode. Statement III is false because electrode potential is an intensive quantity and is not multiplied when a half-reaction is multiplied for electron balance. Coefficients affect stoichiometric electron counts, not the voltage value. Keeping these three ideas separate is essential in cell-potential calculations.
416. A table gives cell notations and \(E^\circ_{\text{cell}}\) values.
| Row | Cell notation | \(E^\circ_{\text{cell}}\) | Decision |
|---|
| P | \( \mathrm{Zn|Zn^{2+}||Cu^{2+}|Cu} \) | \(+1.10\,\mathrm{V}\) | favoured as written |
| Q | \( \mathrm{Cu|Cu^{2+}||Zn^{2+}|Zn} \) | \(-1.10\,\mathrm{V}\) | not favoured as written |
| R | \( \mathrm{Fe|Fe^{2+}||Ag^+|Ag} \) | \(+1.24\,\mathrm{V}\) | favoured as written |
| S | \( \mathrm{Ag|Ag^+||Fe^{2+}|Fe} \) | \(+1.24\,\mathrm{V}\) | favoured as written |
Which row needs correction?
ⓐ. Row P
ⓑ. Row S
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: Row P is supported because zinc oxidation and copper ion reduction give a positive cell potential. Row Q is the reverse of Row P and is not favoured as written, so its negative value and decision are consistent. Row R is supported because iron can be oxidised while silver ions are reduced, giving a positive value. Row S reverses the iron-silver cell but keeps the same positive value, which is incorrect. Reversing the cell direction reverses the sign of \(E^\circ_{\text{cell}}\).
417. A compact cell problem gives \(E^\circ_{\mathrm{X^+/X}}=+0.20\,\mathrm{V}\) and \(E^\circ_{\mathrm{Y^+/Y}}=+0.65\,\mathrm{V}\). The standard cell \( \mathrm{X|X^+||Y^+|Y} \) has:
ⓐ. \(E^\circ_{\text{cell}}=-0.45\,\mathrm{V}\)
ⓑ. \(E^\circ_{\text{cell}}=+0.85\,\mathrm{V}\)
ⓒ. \(E^\circ_{\text{cell}}=+0.45\,\mathrm{V}\)
ⓓ. \(E^\circ_{\text{cell}}=0.00\,\mathrm{V}\)
Correct Answer: \(E^\circ_{\text{cell}}=+0.45\,\mathrm{V}\)
Explanation: \( \textbf{Notation meaning:} \) \( \mathrm{X|X^+} \) is written on the left, so it is the anode side in usual galvanic notation.
\( \textbf{Cathode side:} \) \( \mathrm{Y^+|Y} \) is on the right and represents reduction.
\( \textbf{Use reduction potentials:} \)
\[
E^\circ_{\text{cell}}=E^\circ_{\text{right}}-E^\circ_{\text{left}}
\]
\[
E^\circ_{\text{cell}}=(+0.65)-(+0.20)
\]
\[
E^\circ_{\text{cell}}=+0.45\,\mathrm{V}
\]
\( \textbf{Final answer:} \) The standard cell potential is \(+0.45\,\mathrm{V}\). The values are subtracted because both are tabulated as reduction potentials.
418. The reaction \( \mathrm{X+Y^+\rightarrow X^+ +Y} \) has \(E^\circ_{\text{cell}}=+0.45\,\mathrm{V}\). The oxidised and reduced species are respectively:
ⓐ. \( \mathrm{X} \) and \( \mathrm{Y^+} \)
ⓑ. \( \mathrm{Y^+} \) and \( \mathrm{X} \)
ⓒ. \( \mathrm{X^+} \) and \( \mathrm{Y} \)
ⓓ. \( \mathrm{Y} \) and \( \mathrm{X^+} \)
Correct Answer: \( \mathrm{X} \) and \( \mathrm{Y^+} \)
Explanation: In the reaction, \( \mathrm{X} \) changes from oxidation number \(0\) to \(+1\), so it loses an electron. Loss of electrons is oxidation. \( \mathrm{Y^+} \) changes from \(+1\) to \(0\), so it gains an electron. Gain of electrons is reduction. The positive \(E^\circ_{\text{cell}}\) supports the written direction under standard conditions, but the oxidation and reduction labels still come from the species changes.
419. A cell-potential calculation gives \(E^\circ_{\text{cell}}=(+0.20)-(+0.65)=-0.45\,\mathrm{V}\) for the reaction \( \mathrm{Y+X^+\rightarrow Y^+ +X} \). The conclusion is:
ⓐ. the reaction is favoured because the magnitude is \(0.45\,\mathrm{V}\)
ⓑ. the reaction becomes favoured by doubling all coefficients
ⓒ. the reaction is non-redox because the value is negative
ⓓ. the reaction as written is not favoured under standard conditions
Correct Answer: the reaction as written is not favoured under standard conditions
Explanation: The calculated standard cell potential is negative for the written direction. This means the reverse direction is favoured under standard conditions. The magnitude alone is not enough; the sign must be considered. Doubling coefficients would not change the voltage sign because cell potential is intensive. A negative value does not mean no redox chemistry exists, only that the written direction is not the spontaneous galvanic direction under the stated conditions.
420. The most reliable sequence for solving a standard cell-potential problem is:
ⓐ. add all listed potentials, multiply by coefficients, then choose the larger formula mass
ⓑ. decide the answer from colour change, then ignore electrode notation
ⓒ. choose anode/cathode, use the cell-potential equation, interpret sign
ⓓ. choose the electrode with more atoms as cathode and the other as anode
Correct Answer: choose anode/cathode, use the cell-potential equation, interpret sign
Explanation: A standard cell-potential problem must begin with the redox half-reactions or the cell notation. The cathode is the reduction half-cell, and the anode is the oxidation half-cell. When using standard reduction potentials, the anode reduction potential is subtracted from the cathode reduction potential. The sign of \(E^\circ_{\text{cell}}\) then tells whether the written direction is favoured under standard conditions. Coefficients are used for electron balance but do not multiply electrode potentials.
