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: Start from the free energy:
For any reaction, $\Delta G = \Delta G^{\circ} + RT\ln Q$. Also $\Delta G = -nFE$ and $\Delta G^{\circ} = -nFE^{\circ}$. We use these three links to build the cell equation.

Step 2: Convert to emf:
Substitute the two emf forms: $-nFE = -nFE^{\circ} + RT\ln Q$. Divide every term by $-nF$. This gives $E = E^{\circ} - \dfrac{RT}{nF}\ln Q$.

Step 3: Count the electrons:
The half reactions are $Ni \to Ni^{2+} + 2e^{-}$ and $2Ag^{+} + 2e^{-} \to 2Ag$. The electrons cancel only when the silver half reaction is doubled, so $n = 2$.

Step 4: Write Q:
Solids are left out of $Q$. The product is $Ni^{2+}$ and the reactant is $Ag^{+}$ with coefficient 2. So $Q = [Ni^{2+}]/[Ag^{+}]^{2}$.

Step 5: Compare with the four choices:
Only option 1 has both $2F$ in the denominator and $[Ag^{+}]^{2}$ inside the log, with a minus sign. Options 2 and 3 miss the square. Options 3 and 4 use $F$ instead of $2F$. Option 4 also has a wrong plus sign.

Final Answer:
Using the Nernst equation with $n = 2$ and the squared silver ion term, option 1 is correct.\[ \boxed{E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{2F}\ln\dfrac{[Ni^{2+}]}{[Ag^{+}]^{2}}} \]