Comprehension
Electrochemical cells
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combinations of different half-cells. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not require a salt bridge.
For the cell
\[ \text{Ni(s)} \mid \text{Ni}^{2+}(\text{aq}) \parallel \text{Ag}^+(\text{aq}) \mid \text{Ag(s)} \] The cell reaction is:
\[ \text{Ni(s)} + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Ni}^{2+}(\text{aq}) + 2\text{Ag(s)} \] Nernst equation relates the emf of the cell with standard emf and the concentration of reduced and oxidized species.
Question: 1
The Nernst equation for the given reaction
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
is:
The Nernst equation for the given reaction
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
is:
Show Hint
Remember to square the concentration of \(\text{Ag}^+\) in the denominator of \(Q\) because its stoichiometric coefficient in the balanced cell reaction is \(2\)!
Updated On: Sep 8, 2026
- \(E_\text{cell} = E^\circ_\text{cell} - \dfrac{RT}{2F}\ln \dfrac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}\)
- \(E_\text{cell} = E^\circ_\text{cell} - \dfrac{RT}{2F}\ln \dfrac{[\text{Ni}^{2+}]}{[\text{Ag}^+]}\)
- \(E_\text{cell} = E^\circ_\text{cell} - \dfrac{RT}{F}\ln \dfrac{[\text{Ni}^{2+}]}{[\text{Ag}^+]}\)
- \(E_\text{cell} = E^\circ_\text{cell} + \dfrac{RT}{F}\ln \dfrac{[\text{Ni}^{2+}]}{[\text{Ag}^+]}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
The Nernst equation expresses the electromotive force (\(E_\text{cell}\)) of an electrochemical cell as a function of the standard cell potential (\(E^\circ_\text{cell}\)), temperature (\(T\)), the number of moles of electrons transferred (\(n\)), and the reaction quotient (\(Q\)):
\[ E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q \]
Step 1: Determining the Number of Electrons Transferred (\(n\)):
The overall redox reaction is:
\[ \text{Ni(s)} + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Ni}^{2+}(\text{aq}) + 2\text{Ag(s)} \] Splitting the reaction into its two half-cell reactions:
- Anodic oxidation: \(\text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^-\)
- Cathodic reduction: \(2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag(s)}\)
Two moles of electrons are exchanged during the overall cell process.
Therefore, \(n = 2\).
Step 2: Formulating the Reaction Quotient (\(Q\)):
In writing the expression for the reaction quotient \(Q\), pure solids have unit activity (\(a_{\text{Ni(s)}} = 1\) and \(a_{\text{Ag(s)}} = 1\)).
Taking stoichiometric coefficients into account:
\[ Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} \]
Step 3: Assembling the Nernst Equation:
Substituting \(n = 2\) and the expression for \(Q\) into the general Nernst equation gives:
\[ E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{2F} \ln \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} \] Final Answer:
The correct Nernst equation corresponds to option (A).
The Nernst equation expresses the electromotive force (\(E_\text{cell}\)) of an electrochemical cell as a function of the standard cell potential (\(E^\circ_\text{cell}\)), temperature (\(T\)), the number of moles of electrons transferred (\(n\)), and the reaction quotient (\(Q\)):
\[ E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q \]
Step 1: Determining the Number of Electrons Transferred (\(n\)):
The overall redox reaction is:
\[ \text{Ni(s)} + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Ni}^{2+}(\text{aq}) + 2\text{Ag(s)} \] Splitting the reaction into its two half-cell reactions:
- Anodic oxidation: \(\text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^-\)
- Cathodic reduction: \(2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag(s)}\)
Two moles of electrons are exchanged during the overall cell process.
Therefore, \(n = 2\).
Step 2: Formulating the Reaction Quotient (\(Q\)):
In writing the expression for the reaction quotient \(Q\), pure solids have unit activity (\(a_{\text{Ni(s)}} = 1\) and \(a_{\text{Ag(s)}} = 1\)).
Taking stoichiometric coefficients into account:
\[ Q = \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} \]
Step 3: Assembling the Nernst Equation:
Substituting \(n = 2\) and the expression for \(Q\) into the general Nernst equation gives:
\[ E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{2F} \ln \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} \] Final Answer:
The correct Nernst equation corresponds to option (A).
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Question: 2
The standard electrode potential for the cell
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
is \(0.80\text{ V}\). The standard Gibbs energy for the reaction is:
The standard electrode potential for the cell
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
is \(0.80\text{ V}\). The standard Gibbs energy for the reaction is:
Show Hint
Sign check in electrochemistry:
If \(E^\circ_\text{cell} > 0\), then \(\Delta G^\circ < 0\) (spontaneous reaction).
This allows you to eliminate positive answer choices immediately.
If \(E^\circ_\text{cell} > 0\), then \(\Delta G^\circ < 0\) (spontaneous reaction).
This allows you to eliminate positive answer choices immediately.
Updated On: Sep 8, 2026
- \(-154.379\text{ kJ mol}^{-1}\)
- \(154.379\text{ kJ mol}^{-1}\)
- \(212.2\text{ kJ mol}^{-1}\)
- \(-212.2\text{ kJ mol}^{-1}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
The maximum electrical work obtainable from an electrochemical cell under standard conditions is equal to the decrease in standard Gibbs free energy:
\[ \Delta G^\circ = -n F E^\circ_\text{cell} \] where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant (\(1\text{ F} \approx 96487\text{ C mol}^{-1}\)), and \(E^\circ_\text{cell}\) is the standard electromotive force.
Step 1: Identifying Given Parameters:
- Number of electrons transferred: \(n = 2\)
- Standard Faraday constant: \(F = 96487\text{ C mol}^{-1}\)
- Standard cell potential: \(E^\circ_\text{cell} = 0.80\text{ V}\)
Step 2: Calculating \(\Delta G^\circ\):
Substitute the values into the thermodynamic relation:
\[ \Delta G^\circ = -(2) \times (96487\text{ C mol}^{-1}) \times (0.80\text{ V}) \] Recalling that \(1\text{ C}\cdot\text{V} = 1\text{ J}\):
\[ \Delta G^\circ = -154379.2\text{ J mol}^{-1} \]
Step 3: Converting to Kilojoules per Mole:
Convert the calculated value to \(\text{kJ mol}^{-1}\):
\[ \Delta G^\circ = \frac{-154379.2}{1000}\text{ kJ mol}^{-1} = -154.379\text{ kJ mol}^{-1} \] Because \(E^\circ_\text{cell}\) is positive (\(+0.80\text{ V}\)), the standard Gibbs free energy change \(\Delta G^\circ\) must be negative, reflecting a spontaneous process.
Final Answer:
The standard Gibbs energy for the reaction is \(-154.379\text{ kJ mol}^{-1}\), corresponding to option (A).
The maximum electrical work obtainable from an electrochemical cell under standard conditions is equal to the decrease in standard Gibbs free energy:
\[ \Delta G^\circ = -n F E^\circ_\text{cell} \] where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant (\(1\text{ F} \approx 96487\text{ C mol}^{-1}\)), and \(E^\circ_\text{cell}\) is the standard electromotive force.
Step 1: Identifying Given Parameters:
- Number of electrons transferred: \(n = 2\)
- Standard Faraday constant: \(F = 96487\text{ C mol}^{-1}\)
- Standard cell potential: \(E^\circ_\text{cell} = 0.80\text{ V}\)
Step 2: Calculating \(\Delta G^\circ\):
Substitute the values into the thermodynamic relation:
\[ \Delta G^\circ = -(2) \times (96487\text{ C mol}^{-1}) \times (0.80\text{ V}) \] Recalling that \(1\text{ C}\cdot\text{V} = 1\text{ J}\):
\[ \Delta G^\circ = -154379.2\text{ J mol}^{-1} \]
Step 3: Converting to Kilojoules per Mole:
Convert the calculated value to \(\text{kJ mol}^{-1}\):
\[ \Delta G^\circ = \frac{-154379.2}{1000}\text{ kJ mol}^{-1} = -154.379\text{ kJ mol}^{-1} \] Because \(E^\circ_\text{cell}\) is positive (\(+0.80\text{ V}\)), the standard Gibbs free energy change \(\Delta G^\circ\) must be negative, reflecting a spontaneous process.
Final Answer:
The standard Gibbs energy for the reaction is \(-154.379\text{ kJ mol}^{-1}\), corresponding to option (A).
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Question: 3
The relationship between the equilibrium constant of the reaction and the standard electrode potential of the cell in which that reaction takes place is given by
The relationship between the equilibrium constant of the reaction and the standard electrode potential of the cell in which that reaction takes place is given by
Show Hint
At \(298\text{ K}\), substituting numerical values into this equation gives the useful form:
\[ E^\circ_\text{cell} = \frac{0.0591}{n}\log K_c \]
\[ E^\circ_\text{cell} = \frac{0.0591}{n}\log K_c \]
Updated On: Sep 7, 2026
- \(E^\circ_\text{cell} = \dfrac{RT}{2.303 \times nF}\log K_c\)
- \(E^\circ_\text{cell} = \dfrac{2.303RT}{nF}\log K_c\)
- \(E^\circ_\text{cell} = \dfrac{2.303RT}{nF}\ln K_c\)
- \(E^\circ_\text{cell} = 2.303RT\ln K_c\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
At dynamic chemical equilibrium, the cell electromotive force reaches zero (\(E_\text{cell} = 0\)) and the reaction quotient becomes equal to the equilibrium constant (\(Q = K_c\)).
Substituting these boundary conditions into the Nernst equation provides the thermodynamic link between standard potential and the equilibrium constant.
Step 1: Applying Equilibrium Conditions to the Nernst Equation:
The general Nernst equation is:
\[ E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q \] Setting \(E_\text{cell} = 0\) and \(Q = K_c\):
\[ 0 = E^\circ_\text{cell} - \frac{RT}{nF} \ln K_c \] Rearranging gives:
\[ E^\circ_\text{cell} = \frac{RT}{nF} \ln K_c \]
Step 2: Converting Natural Logarithm to Base-10 Logarithm:
Using the identity relating natural and common logarithms:
\[ \ln K_c = 2.303 \log_{10} K_c \] Substituting this into the expression gives:
\[ E^\circ_\text{cell} = \frac{2.303 RT}{nF} \log K_c \]
Step 3: Analyzing the Given Choices:
- Option (B) correctly includes the factor \(2.303\) alongside the common logarithm \(\log K_c\).
- Option (C) incorrectly combines \(2.303\) with the natural logarithm \(\ln K_c\).
- Option (A) has the \(2.303\) factor in the denominator.
- Option (D) omits the \(nF\) term entirely.
Final Answer:
The correct relationship is \(E^\circ_\text{cell} = \dfrac{2.303RT}{nF}\log K_c\), corresponding to option (B).
At dynamic chemical equilibrium, the cell electromotive force reaches zero (\(E_\text{cell} = 0\)) and the reaction quotient becomes equal to the equilibrium constant (\(Q = K_c\)).
Substituting these boundary conditions into the Nernst equation provides the thermodynamic link between standard potential and the equilibrium constant.
Step 1: Applying Equilibrium Conditions to the Nernst Equation:
The general Nernst equation is:
\[ E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q \] Setting \(E_\text{cell} = 0\) and \(Q = K_c\):
\[ 0 = E^\circ_\text{cell} - \frac{RT}{nF} \ln K_c \] Rearranging gives:
\[ E^\circ_\text{cell} = \frac{RT}{nF} \ln K_c \]
Step 2: Converting Natural Logarithm to Base-10 Logarithm:
Using the identity relating natural and common logarithms:
\[ \ln K_c = 2.303 \log_{10} K_c \] Substituting this into the expression gives:
\[ E^\circ_\text{cell} = \frac{2.303 RT}{nF} \log K_c \]
Step 3: Analyzing the Given Choices:
- Option (B) correctly includes the factor \(2.303\) alongside the common logarithm \(\log K_c\).
- Option (C) incorrectly combines \(2.303\) with the natural logarithm \(\ln K_c\).
- Option (A) has the \(2.303\) factor in the denominator.
- Option (D) omits the \(nF\) term entirely.
Final Answer:
The correct relationship is \(E^\circ_\text{cell} = \dfrac{2.303RT}{nF}\log K_c\), corresponding to option (B).
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Question: 4
In the given cell,
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
which of the given species is the reducing agent?
In the given cell,
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
which of the given species is the reducing agent?
Show Hint
Remember:
- Reducing agent = Reactant that undergoes oxidation (loses \(e^-\)).
- Oxidizing agent = Reactant that undergoes reduction (gains \(e^-\)).
Products cannot be the agents for the forward reaction!
- Reducing agent = Reactant that undergoes oxidation (loses \(e^-\)).
- Oxidizing agent = Reactant that undergoes reduction (gains \(e^-\)).
Products cannot be the agents for the forward reaction!
Updated On: Sep 8, 2026
- \(\text{Ni}^{2+}\text{(aq)}\)
- \(\text{Ni(s)}\)
- \(\text{Ag}^+\text{(aq)}\)
- \(\text{Ag(s)}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
In a redox reaction:
- The species that loses electrons undergoes oxidation and acts as the reducing agent (reductant).
- The species that gains electrons undergoes reduction and acts as the oxidizing agent (oxidant).
Step 1: Tracking Oxidation States:
Let us evaluate the changes in oxidation number for all elements in the cell reaction:
\[ \overset{0}{\text{Ni}}(\text{s}) + 2\overset{+1}{\text{Ag}}^+(\text{aq}) \rightarrow \overset{+2}{\text{Ni}}^{2+}(\text{aq}) + 2\overset{0}{\text{Ag}}(\text{s}) \] 1. Nickel: The oxidation state of solid nickel increases from \(0\) in \(\text{Ni(s)}\) to \(+2\) in \(\text{Ni}^{2+}(\text{aq})\).
This increase in oxidation number indicates that nickel loses two electrons:
\[ \text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^- \quad (\text{Oxidation}) \] 2. Silver: The oxidation state of silver decreases from \(+1\) in \(\text{Ag}^+(\text{aq})\) to \(0\) in \(\text{Ag(s)}\).
This decrease in oxidation number indicates that silver ions gain electrons:
\[ 2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag(s)} \quad (\text{Reduction}) \]
Step 2: Identifying the Reducing Agent:
Because \(\text{Ni(s)}\) is oxidized by losing electrons to \(\text{Ag}^+\), it causes the reduction of silver ions.
Therefore, metallic nickel, \(\text{Ni(s)}\), serves as the reducing agent.
\(\text{Ag}^+(\text{aq})\) is the oxidizing agent, while \(\text{Ni}^{2+}(\text{aq})\) and \(\text{Ag(s)}\) are reaction products.
Final Answer:
The reducing agent is \(\text{Ni(s)}\), which corresponds to option (B).
In a redox reaction:
- The species that loses electrons undergoes oxidation and acts as the reducing agent (reductant).
- The species that gains electrons undergoes reduction and acts as the oxidizing agent (oxidant).
Step 1: Tracking Oxidation States:
Let us evaluate the changes in oxidation number for all elements in the cell reaction:
\[ \overset{0}{\text{Ni}}(\text{s}) + 2\overset{+1}{\text{Ag}}^+(\text{aq}) \rightarrow \overset{+2}{\text{Ni}}^{2+}(\text{aq}) + 2\overset{0}{\text{Ag}}(\text{s}) \] 1. Nickel: The oxidation state of solid nickel increases from \(0\) in \(\text{Ni(s)}\) to \(+2\) in \(\text{Ni}^{2+}(\text{aq})\).
This increase in oxidation number indicates that nickel loses two electrons:
\[ \text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^- \quad (\text{Oxidation}) \] 2. Silver: The oxidation state of silver decreases from \(+1\) in \(\text{Ag}^+(\text{aq})\) to \(0\) in \(\text{Ag(s)}\).
This decrease in oxidation number indicates that silver ions gain electrons:
\[ 2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag(s)} \quad (\text{Reduction}) \]
Step 2: Identifying the Reducing Agent:
Because \(\text{Ni(s)}\) is oxidized by losing electrons to \(\text{Ag}^+\), it causes the reduction of silver ions.
Therefore, metallic nickel, \(\text{Ni(s)}\), serves as the reducing agent.
\(\text{Ag}^+(\text{aq})\) is the oxidizing agent, while \(\text{Ni}^{2+}(\text{aq})\) and \(\text{Ag(s)}\) are reaction products.
Final Answer:
The reducing agent is \(\text{Ni(s)}\), which corresponds to option (B).
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Question: 5
Which of the following is the anodic half-cell reaction?
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
Which of the following is the anodic half-cell reaction?
\[ \text{Ni(s)} + 2\text{Ag}^{+}\text{(aq)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2\text{Ag(s)} \]
Show Hint
Standard mnemonic:
- Anode = Oxidation (An Ox).
- Cathode = Reduction (Red Cat).
In galvanic cells, the anode is written on the left.
- Anode = Oxidation (An Ox).
- Cathode = Reduction (Red Cat).
In galvanic cells, the anode is written on the left.
Updated On: Sep 8, 2026
- \(\text{Ni(s)} \rightarrow \text{Ni}^{2+}\text{(aq)} + 2e^-\)
- \(\text{Ni}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Ni(s)}\)
- \(\text{Ag}^+\text{(aq)} + e^- \rightarrow \text{Ag(s)}\)
- \(\text{Ag(s)} \rightarrow \text{Ag}^+\text{(aq)} + e^-\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
In all electrochemical and galvanic cells, the electrode at which oxidation (loss of electrons) occurs is designated as the anode.
The electrode at which reduction (gain of electrons) occurs is designated as the cathode.
Step 1: Analyzing Cell Notation and Half-Reactions:
The cell notation given in the comprehension is:
\[ \text{Ni(s)} \mid \text{Ni}^{2+}(\text{aq}) \parallel \text{Ag}^+(\text{aq}) \mid \text{Ag(s)} \] By IUPAC convention:
- The half-cell written on the left-hand side is the anode.
- The half-cell written on the right-hand side is the cathode.
Step 2: Formulating the Oxidation Reaction at the Anode:
At the nickel electrode (anode), metallic nickel loses two electrons to enter the solution as hydrated nickel(II) cations:
\[ \text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^- \] This is the anodic half-cell reaction.
Step 3: Formulating the Reduction Reaction at the Cathode:
At the silver electrode (cathode), silver ions in solution accept electrons from the electrode to deposit as metallic silver:
\[ \text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag(s)} \] This represents the cathodic half-cell reaction.
Final Answer:
The anodic half-cell reaction is \(\text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^-\), which corresponds to option (A).
In all electrochemical and galvanic cells, the electrode at which oxidation (loss of electrons) occurs is designated as the anode.
The electrode at which reduction (gain of electrons) occurs is designated as the cathode.
Step 1: Analyzing Cell Notation and Half-Reactions:
The cell notation given in the comprehension is:
\[ \text{Ni(s)} \mid \text{Ni}^{2+}(\text{aq}) \parallel \text{Ag}^+(\text{aq}) \mid \text{Ag(s)} \] By IUPAC convention:
- The half-cell written on the left-hand side is the anode.
- The half-cell written on the right-hand side is the cathode.
Step 2: Formulating the Oxidation Reaction at the Anode:
At the nickel electrode (anode), metallic nickel loses two electrons to enter the solution as hydrated nickel(II) cations:
\[ \text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^- \] This is the anodic half-cell reaction.
Step 3: Formulating the Reduction Reaction at the Cathode:
At the silver electrode (cathode), silver ions in solution accept electrons from the electrode to deposit as metallic silver:
\[ \text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag(s)} \] This represents the cathodic half-cell reaction.
Final Answer:
The anodic half-cell reaction is \(\text{Ni(s)} \rightarrow \text{Ni}^{2+}(\text{aq}) + 2e^-\), which corresponds to option (A).
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