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JEE Advance - Chemistry (2010 - Paper 1 Offline - No. 10)

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :
M(s) | M+ (aq ; 0.05 molar) || M+ (aq ; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV.

For the above cell :
Ecell < 0 ; $$\Delta G > 0$$
Ecell > 0 ; $$\Delta G < 0$$
Ecell < 0 ; $$\Delta G^o > 0$$
Ecell > 0 ; $$\Delta G^o > 0$$

Explanation

To determine the right option for the given concentration cell, we must first understand the concepts of cell potential, Gibbs free energy, and the relationship between these quantities.

The cell potential, $$ E_{cell} $$, for a concentration cell like the one described is directly related to the concentrations of the ions on both sides of the cell. It’s calculated using the Nernst equation:

$$ E_{cell} = E^o - \frac{RT}{nF} \ln \frac{[M^+](\text{low concentration})}{[M^+](\text{high concentration})} $$

Since the given cell involves the same metal on both sides in its standard state, the standard cell potential, $$ E^o $$, is zero. Therefore, the equation simplifies to:

$$ E_{cell} = - \frac{RT}{nF} \ln \frac{0.05}{1} $$

Given that $$ R $$ (the ideal gas constant) is approximately 8.314 J/mol·K, $$ F $$ (the Faraday constant) is about 96485 C/mol, $$ T $$ is the temperature in Kelvin (assuming standard temperature of 298 K), and $$ n $$ (the number of moles of electrons transferred per mole of reaction) is 1, the equation further simplifies to:

$$ E_{cell} = - \frac{8.314 \times 298}{96485} \ln \left(\frac{0.05}{1}\right) $$

Calculate the ln term:

$$ \ln \left(\frac{0.05}{1}\right) = \ln(0.05) \approx -2.9957 $$

Then the equation is:

$$ E_{cell} = - \frac{2476.422}{96485} \times -2.9957 \approx 0.077 \text{ volts} = 77 \text{ mV} $$

Here, the calculated $$ E_{cell} > 0 $$, which matches the given cell potential of 70 mV. It differs slightly due to rounding and exact values used in constants. This cell potential being positive indicates spontaneous reaction (tendency to go from high to low concentration).

We now apply this to the relationship between the cell potential and Gibbs free energy, which is given by:

$$ \Delta G = -nFE_{cell} $$

Since $$ E_{cell} > 0 $$, $$ \Delta G < 0 $$ which indicates that the process is thermodynamically favorable (spontaneous).

Looking at the provided options:

  • Option A: Ecell < 0 ; $$\Delta G > 0$$ (Incorrect; as $$ E_{cell} $$ is positive and $$ \Delta G $$ is negative)
  • Option B: Ecell > 0 ; $$\Delta G < 0$$ (Correct; matches our calculation)
  • Option C: Ecell < 0 ; $$\Delta G^o > 0$$ (Incorrect)
  • Option D: Ecell > 0 ; $$\Delta G^o > 0$$ (Incorrect, $$ \Delta G^o $$ not directly relevant here, but centralized on the standard conditions which were zero)

Thus, the correct answer is Option B.

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