Questions: Which of the following statements is true concerning the hypothetical electrochemical cell depicted below? (M1 and M2 are differing metals) M1M^2+(aq) M^+(aq)M2 M^2+(aq)+2e^- M1(s) ; E°=-2.91 V M^2+(aq)+e^- M2(s) ; E°=-2.98 V The cell reaction is spontaneous with a standard cell potential of 0.07 V. The cell reaction is nonspontaneous with a standard cell potential of -0.07 V. The cell reaction is spontaneous with a standard cell potential of 5.89 V. The cell is at equilibrium. The cell reaction is nonspontaneous with a standard cell potential of -5.89 V.

Which of the following statements is true concerning the hypothetical electrochemical cell depicted below? (M1 and M2 are differing metals)

M1M^2+(aq)  M^+(aq)M2
M^2+(aq)+2e^- M1(s) ; E°=-2.91 V
M^2+(aq)+e^- M2(s) ; E°=-2.98 V

The cell reaction is spontaneous with a standard cell potential of 0.07 V.
The cell reaction is nonspontaneous with a standard cell potential of -0.07 V.
The cell reaction is spontaneous with a standard cell potential of 5.89 V.
The cell is at equilibrium.
The cell reaction is nonspontaneous with a standard cell potential of -5.89 V.
Transcript text: Which of the following statements is true concerning the hypothetical electrochemical cell depicted below? (M1 and M2 are differing metals) \[ \begin{array}{l} \mathrm{M} 1\left|\mathrm{M}^{2+}(a q) \| \mathrm{M}^{+}(a q)\right| \mathrm{M} 2 \\ \mathrm{M}^{2+}(a q)+2 \mathrm{e}^{-} \mathrm{M} 1(s) ; E^{\circ}=-2.91 \mathrm{~V} \\ \mathrm{M}^{2+}(a q)+\mathrm{e}^{-} \mathrm{M} 2(s) ; E^{\circ}=-2.98 \mathrm{~V} \end{array} \] The cell reaction is spontaneous with a standard cell potential of 0.07 V . The cell reaction is nonspontaneous with a standard cell potential of -0.07 V . The cell reaction is spontaneous with a standard cell potential of 5.89 V . The cell is at equilibrium. The cell reaction is nonspontaneous with a standard cell potential of -5.89 V .
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Solution

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Solution Steps

Step 1: Identify the Half-Reactions and Their Potentials

The given half-reactions and their standard electrode potentials are:

  1. \(\mathrm{M}^{2+}(aq) + 2\mathrm{e}^{-} \rightarrow \mathrm{M}1(s)\); \(E^{\circ} = -2.91 \, \text{V}\)
  2. \(\mathrm{M}^{+}(aq) + \mathrm{e}^{-} \rightarrow \mathrm{M}2(s)\); \(E^{\circ} = -2.98 \, \text{V}\)
Step 2: Determine the Cell Reaction

The cell is set up as \(\mathrm{M}1|\mathrm{M}^{2+}(aq) \| \mathrm{M}^{+}(aq)|\mathrm{M}2\). This implies:

  • The anode (oxidation) reaction is: \(\mathrm{M}1(s) \rightarrow \mathrm{M}^{2+}(aq) + 2\mathrm{e}^{-}\)
  • The cathode (reduction) reaction is: \(\mathrm{M}^{+}(aq) + \mathrm{e}^{-} \rightarrow \mathrm{M}2(s)\)
Step 3: Calculate the Standard Cell Potential

The standard cell potential \(E^{\circ}_{\text{cell}}\) is calculated as:

\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]

Substituting the given values:

\[ E^{\circ}_{\text{cell}} = (-2.98 \, \text{V}) - (-2.91 \, \text{V}) = -2.98 \, \text{V} + 2.91 \, \text{V} = -0.07 \, \text{V} \]

Step 4: Determine the Spontaneity of the Reaction

A negative standard cell potential (\(E^{\circ}_{\text{cell}} < 0\)) indicates that the cell reaction is nonspontaneous under standard conditions.

Final Answer

The cell reaction is nonspontaneous with a standard cell potential of \(-0.07 \, \text{V}\).

\[ \boxed{\text{The cell reaction is nonspontaneous with a standard cell potential of } -0.07 \, \text{V}.} \]

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