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 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

• A. \(E^{\circ}_{cell} = \dfrac{RT}{2.303 \times nF}\log K_c\)
• B. \(E^{\circ}_{cell} = \dfrac{2.303RT}{nF}\log K_c\)
• C. \(E^{\circ}_{cell} = \dfrac{2.303RT}{nF}\ln K_c\)
• D. \(E^{\circ}_{cell} = 2.303RT\ln K_c\)

Hint:

At equilibrium E = 0 in the Nernst equation; then convert ln to 2.303 log.

Answer 1

The Correct Option is B

Step 1: Think about the units:
$E^{\circ}$ is in volts. $RT$ is in joules per mole, and $nF$ is in coulombs per mole, and J/C is V. So the right formula must carry $RT/nF$. Option 4 lacks $nF$, so it is out.

Step 2: Link with Gibbs energy:
$\Delta G^{\circ} = -nFE^{\circ}$ and $\Delta G^{\circ} = -RT\ln K_c$. Equate both: $nFE^{\circ} = RT\ln K_c$.

Step 3: Solve for E:
\[ E^{\circ}_{cell} = \dfrac{RT}{nF}\ln K_c \]

Step 4: Change ln to log:
Replace $\ln K_c$ with $2.303\log K_c$. We get $E^{\circ}_{cell} = \dfrac{2.303RT}{nF}\log K_c$.

Step 5: Match:
Only option 2 has 2.303 in the numerator together with $\log$. Option 3 mixes 2.303 with $\ln$, and option 1 puts 2.303 in the denominator.

Final Answer:
Combining delta G = -nFE and delta G = -RT ln K gives E = (2.303RT/nF) log K, option 2.\[ \boxed{E^{\circ}_{cell} = \dfrac{2.303RT}{nF}\log K_c} \]