JEE Challenger
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ChemMedAdv 2010 • Paper 1

Determine Cell Potential Sign and Free Energy Change for Cell

Comprehension Passage

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\text{M} is:
M(s)∣M+(aq; 0.05 molar)∣∣M+(aq; 1 molar)∣M(s)\text{M(s)} \mid \text{M}^+(\text{aq; } 0.05\text{ molar}) \mid\mid \text{M}^+(\text{aq; } 1\text{ molar}) \mid \text{M(s)}
For the above electrolytic cell the magnitude of the cell potential ∣Ecell∣=70 mV|E_{\text{cell}}| = 70\text{ mV}.

For the above cell

Options

A

Ecell<0;ΔG>0E_{\text{cell}} < 0; \Delta G > 0

B

Ecell>0;ΔG<0E_{\text{cell}} > 0; \Delta G < 0

Correct
C

Ecell<0;ΔG0>0E_{\text{cell}} < 0; \Delta G^0 > 0

D

Ecell>0;ΔG0<0E_{\text{cell}} > 0; \Delta G^0 < 0

Topics & Concepts

ElectrochemistryElectrochemical CellGibbs Free Energy

Step-by-Step Solution

To determine the signs of the cell potential (EcellE_{\text{cell}}) and the Gibbs free energy change (ΔG\Delta G) for the given concentration cell, we write down the respective half-cell reactions and the overall cell process.

  1. Cell Notation and Half-Cell Reactions:
    The representation of the concentration cell is:
    M(s)∣M+(aq; 0.05 M)∣∣M+(aq; 1 M)∣M(s)\text{M(s)} \mid \text{M}^+(\text{aq; } 0.05\text{ M}) \mid\mid \text{M}^+(\text{aq; } 1\text{ M}) \mid \text{M(s)}

    • Anode (Oxidation):
      M(s)→M+(aq; C1=0.05 M)+e−\text{M(s)} \rightarrow \text{M}^+(\text{aq; } C_1 = 0.05\text{ M}) + \text{e}^-

    • Cathode (Reduction):
      M+(aq; C2=1 M)+e−→M(s)\text{M}^+(\text{aq; } C_2 = 1\text{ M}) + \text{e}^- \rightarrow \text{M(s)}

    • Overall Cell Reaction:
      M+(aq; 1 M)⇌M+(aq; 0.05 M)\text{M}^+(\text{aq; } 1\text{ M}) \rightleftharpoons \text{M}^+(\text{aq; } 0.05\text{ M})

  2. Calculation of Cell Potential (EcellE_{\text{cell}}):
    For any concentration cell consisting of the same electrodes, the standard cell potential is zero:
    Ecell∘=0 VE^\circ_{\text{cell}} = 0\text{ V}

    Applying the Nernst equation at temperature TT:
    Ecell=Ecell∘−RTnFln⁡([M+]anode[M+]cathode)E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln \left( \frac{[\text{M}^+]_{\text{anode}}}{[\text{M}^+]_{\text{cathode}}} \right)

    Substitute the given concentration values (n=1n = 1):
    Ecell=0−RTFln⁡(0.051)=−RTFln⁡(0.05)E_{\text{cell}} = 0 - \frac{RT}{F} \ln \left( \frac{0.05}{1} \right) = -\frac{RT}{F} \ln(0.05)

    Since 0.05<10.05 < 1, the natural logarithm ln⁡(0.05)\ln(0.05) is negative (ln⁡(0.05)≈−2.996\ln(0.05) \approx -2.996).
    Thus, the negative signs cancel out, giving:
    Ecell>0E_{\text{cell}} > 0

  3. Determination of Gibbs Free Energy Change (ΔG\Delta G):
    The relation between the Gibbs free energy change and the cell potential is given by:
    ΔG=−nFEcell\Delta G = -nFE_{\text{cell}}

    Since n>0n > 0, F>0F > 0, and Ecell>0E_{\text{cell}} > 0, it follows directly that:
    ΔG<0\Delta G < 0

Therefore, for the given cell, Ecell>0E_{\text{cell}} > 0 and ΔG<0\Delta G < 0.

Correct Answer: B

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