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:

• A. -154.379 kJ mol\(^{-1}\)
• B. 154.379 kJ mol\(^{-1}\)
• C. 212.2 kJ mol\(^{-1}\)
• D. -212.2 kJ mol\(^{-1}\)

Hint:

Use delta G = -nFE with n = 2 and keep the negative sign.

Answer 1

The Correct Option is A

Step 1: Decide the sign first:
The cell emf is positive (0.80 V). For a spontaneous reaction $\Delta G^{\circ}$ is negative. This removes options 2 and 3 straight away.

Step 2: Use the relation:
$\Delta G^{\circ} = -nFE^{\circ}$ with $n = 2$, since two electrons move from Ni to the silver ions.

Step 3: Do the arithmetic in joules:
\[ 2 \times 96487 = 192974 \]
\[ 192974 \times 0.80 = 154379.2\ \text{J mol}^{-1} \]

Step 4: Change to kJ:
Divide by 1000 to get 154.379 kJ mol$^{-1}$. Keep the negative sign, so $\Delta G^{\circ} = -154.379$ kJ mol$^{-1}$.

Step 5: Rule out option (4):
$-212.2$ kJ would need $E^{\circ} \approx 1.10$ V for $n = 2$. The given value is 0.80 V, so option 4 does not fit.

Final Answer:
With $n = 2$ and $E^{\circ} = 0.80$ V, the standard Gibbs energy is -154.379 kJ per mole.\[ \boxed{\Delta G^{\circ} = -154.379\ \text{kJ mol}^{-1}} \]