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: Understanding the Concept:
At equilibrium the cell emf becomes zero, because the cell can do no more work. Putting this in the Nernst equation links the standard emf to the equilibrium constant.

Step 2: Key Formula or Approach:
Start from \(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{nF}\ln Q\). At equilibrium \(E_{cell} = 0\) and \(Q = K_c\). Then use \(\ln x = 2.303\log x\).

Step 3: Derive:
\[ 0 = E^{\circ}_{cell} - \dfrac{RT}{nF}\ln K_c \]
\[ E^{\circ}_{cell} = \dfrac{RT}{nF}\ln K_c = \dfrac{2.303RT}{nF}\log K_c \]

Step 4: Check option (1):
It has 2.303 in the denominator. That is the wrong way round, because converting \(\ln\) to \(\log\) multiplies by 2.303. So option 1 is WRONG.

Step 5: Check option (2):
It is \(\dfrac{2.303RT}{nF}\log K_c\), the same as our result. So option 2 is CORRECT.

Step 6: Check option (3):
It keeps 2.303 but uses \(\ln K_c\). The factor 2.303 is only there when we use \(\log_{10}\). With \(\ln\) it should be absent. So option 3 is WRONG.

Step 7: Check option (4):
It leaves out \(nF\) from the denominator, so the units do not work out. Volts need \(RT/nF\). So option 4 is WRONG.

Final Answer:
The correct relation is \(E^{\circ}_{cell} = \dfrac{2.303RT}{nF}\log K_c\), option 2.\[ \boxed{E^{\circ}_{cell} = \dfrac{2.303RT}{nF}\log K_c} \]