Which from following statements is true regarding the cell emf at 298 K for \( \ominus Ni_{(s)} | Ni^{+2}(0.01M) || Ag^{+}(0.01M) | Ag_{(s)} \oplus \)?
Show Hint
$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q$. Pay attention to the coefficients in the balanced equation.
Step 1: Understanding the Question:
We need to determine the relationship between the cell potential (\( E_{cell} \)) and the standard cell potential (\( E^{\circ}_{cell} \)) for the given concentration cell conditions using the Nernst equation. Step 2: Key Formula or Approach:
Cell reaction: \( Ni_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Ni^{2+}_{(aq)} + 2Ag_{(s)} \)
Number of electrons transferred, \( n = 2 \).
Nernst Equation:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^{2}} \] Step 3: Detailed Explanation:
Substitute the given concentrations: \( [Ni^{2+}] = 0.01\text{ M} \), \( [Ag^{+}] = 0.01\text{ M} \).
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log \frac{0.01}{(0.01)^{2}} \]
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log \frac{10^{-2}}{10^{-4}} \]
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log(10^{2}) \]
Since \( \log(10^{2}) = 2 \):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \times 2 \]
\[ E_{cell} = E^{\circ}_{cell} - 0.0592\text{ V} \]
This means \( E_{cell} \) is less than \( E^{\circ}_{cell} \) by 0.0592 V. Step 4: Final Answer:
The true statement is that the emf is less than \( E^{\circ}_{cell} \) by 0.0592 V.