Question:
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
\(\mathrm{Ni(s) \mid Ni^{2+}(aq) \parallel Ag^{+}(aq) \mid Ag}\)
The cell reaction is \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\).
Nernst equation relates the emf of the cell with standard emf and the concentration of reduced and oxidized species.
The standard electrode potential for the cell
\(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\)
is 0.80 V. The standard Gibbs energy for the reaction is:
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
\(\mathrm{Ni(s) \mid Ni^{2+}(aq) \parallel Ag^{+}(aq) \mid Ag}\)
The cell reaction is \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\).
Nernst equation relates the emf of the cell with standard emf and the concentration of reduced and oxidized species.
The standard electrode potential for the cell
\(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\)
is 0.80 V. The standard Gibbs energy for the reaction is:
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
\(\mathrm{Ni(s) \mid Ni^{2+}(aq) \parallel Ag^{+}(aq) \mid Ag}\)
The cell reaction is \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\).
Nernst equation relates the emf of the cell with standard emf and the concentration of reduced and oxidized species.
The standard electrode potential for the cell
\(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\)
is 0.80 V. The standard Gibbs energy for the reaction is:
Show Hint
Use delta G = -nFE with n = 2 and keep the negative sign.
Updated On: Oct 1, 2026
- -154.379 kJ mol\(^{-1}\)
- 154.379 kJ mol\(^{-1}\)
- 212.2 kJ mol\(^{-1}\)
- -212.2 kJ mol\(^{-1}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The standard Gibbs energy change of a cell reaction is linked to the standard emf. A positive emf means a spontaneous reaction, so \(\Delta G^{\circ}\) must come out negative.
Step 2: Key Formula or Approach:
\(\Delta G^{\circ} = -nFE^{\circ}_{cell}\). Here \(n\) is the number of electrons transferred and \(F = 96487\) C mol\(^{-1}\) is the Faraday constant.
Step 3: Find n:
In \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\), Ni gives up 2 electrons and two \(Ag^{+}\) ions take 2 electrons. So \(n = 2\).
Step 4: Calculate:
\[ \Delta G^{\circ} = -(2)(96487\ \text{C mol}^{-1})(0.80\ \text{V}) \]
\[ \Delta G^{\circ} = -154379.2\ \text{J mol}^{-1} = -154.379\ \text{kJ mol}^{-1} \]
Step 5: Check option (1):
The value \(-154.379\) kJ mol\(^{-1}\) is exactly what we got. So option 1 is CORRECT.
Step 6: Check option (2):
The magnitude is right but the sign is positive. A positive \(\Delta G^{\circ}\) would mean a non-spontaneous reaction, which does not fit a positive \(E^{\circ}\). So option 2 is WRONG.
Step 7: Check options (3) and (4):
The number 212.2 does not come from \(nFE^{\circ}\) with \(n = 2\) and \(E^{\circ} = 0.80\) V. It would need \(E^{\circ}\) of about 1.10 V, which is the Daniell cell value. So options 3 and 4 are WRONG.
Final Answer:
\(\Delta G^{\circ} = -nFE^{\circ} = -154.379\) kJ mol\(^{-1}\), which is option 1.\[ \boxed{\Delta G^{\circ} = -154.379\ \text{kJ mol}^{-1}} \]
The standard Gibbs energy change of a cell reaction is linked to the standard emf. A positive emf means a spontaneous reaction, so \(\Delta G^{\circ}\) must come out negative.
Step 2: Key Formula or Approach:
\(\Delta G^{\circ} = -nFE^{\circ}_{cell}\). Here \(n\) is the number of electrons transferred and \(F = 96487\) C mol\(^{-1}\) is the Faraday constant.
Step 3: Find n:
In \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\), Ni gives up 2 electrons and two \(Ag^{+}\) ions take 2 electrons. So \(n = 2\).
Step 4: Calculate:
\[ \Delta G^{\circ} = -(2)(96487\ \text{C mol}^{-1})(0.80\ \text{V}) \]
\[ \Delta G^{\circ} = -154379.2\ \text{J mol}^{-1} = -154.379\ \text{kJ mol}^{-1} \]
Step 5: Check option (1):
The value \(-154.379\) kJ mol\(^{-1}\) is exactly what we got. So option 1 is CORRECT.
Step 6: Check option (2):
The magnitude is right but the sign is positive. A positive \(\Delta G^{\circ}\) would mean a non-spontaneous reaction, which does not fit a positive \(E^{\circ}\). So option 2 is WRONG.
Step 7: Check options (3) and (4):
The number 212.2 does not come from \(nFE^{\circ}\) with \(n = 2\) and \(E^{\circ} = 0.80\) V. It would need \(E^{\circ}\) of about 1.10 V, which is the Daniell cell value. So options 3 and 4 are WRONG.
Final Answer:
\(\Delta G^{\circ} = -nFE^{\circ} = -154.379\) kJ mol\(^{-1}\), which is option 1.\[ \boxed{\Delta G^{\circ} = -154.379\ \text{kJ mol}^{-1}} \]
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