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 Nernst equation for the given reaction
\(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\); is

• A. \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{2F} \ln \dfrac{[Ni^{2+}]}{[Ag^{+}]^{2}}\)
• B. \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{2F} \ln \dfrac{[Ni^{2+}]}{[Ag^{+}]}\)
• C. \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{F} \ln \dfrac{[Ni^{2+}]}{[Ag^{+}]}\)
• D. \(E_{cell} = E^{\circ}_{cell} + \dfrac{RT}{F} \ln \dfrac{[Ni^{2+}]}{[Ag^{+}]}\)

Hint:

Count the electrons transferred (n = 2) and raise each concentration in Q to its coefficient.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Nernst equation gives the emf of a cell when the ions are not at 1 M. It links the actual emf \(E_{cell}\) to the standard emf \(E^{\circ}_{cell}\) through the reaction quotient \(Q\). We need to write \(Q\) correctly for the given reaction and find the number of electrons \(n\).

Step 2: Key Formula or Approach:
The general form is \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{nF}\ln Q\). Here \(Q\) is products over reactants, each raised to its stoichiometric coefficient. Pure solids such as Ni(s) and Ag(s) have activity 1, so they do not appear in \(Q\).

Step 3: Find n and Q:
In the reaction \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\), Ni loses 2 electrons: \(Ni \to Ni^{2+} + 2e^{-}\). Two \(Ag^{+}\) ions gain one electron each. So \(n = 2\).
The reaction quotient is \(Q = \dfrac{[Ni^{2+}]}{[Ag^{+}]^{2}}\), because the coefficient of \(Ag^{+}\) is 2.

Step 4: Check option (1):
Putting \(n = 2\) and the \(Q\) above gives \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{2F}\ln\dfrac{[Ni^{2+}]}{[Ag^{+}]^{2}}\). This matches option 1 exactly. So option 1 is CORRECT.

Step 5: Check option (2):
This has \(n = 2\) but the concentration of \(Ag^{+}\) is not squared. The coefficient 2 of \(Ag^{+}\) must appear as a power in \(Q\). So option 2 is WRONG.

Step 6: Check option (3):
This uses \(n = 1\) and also leaves out the square on \([Ag^{+}]\). The reaction transfers 2 electrons in total, so \(n\) is 2 and not 1. So option 3 is WRONG.

Step 7: Check option (4):
This has a plus sign before the log term, and also uses \(n = 1\) with no square. The Nernst equation always has a minus sign in front of \(\ln Q\). So option 4 is WRONG.

Final Answer:
The Nernst equation for this reaction has \(n = 2\) and \(Q = [Ni^{2+}]/[Ag^{+}]^{2}\). This is option 1.\[ \boxed{E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{2F}\ln\dfrac{[Ni^{2+}]}{[Ag^{+}]^{2}}} \]