JEE Challenger
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ChemMedAdv 2016 • Paper 2

Determine Exponential Parameter in Electrochemical Cell Nernst Equation

For the given galvanic cell operating at 298 K298\text{ K}, Pt(s)∣H2(g,1 bar)∣H+(aq,1 M)∥M4+(aq),M2+(aq)∣Pt(s)\text{Pt}(s) \mid \text{H}_2(g, 1\text{ bar}) \mid \text{H}^+(aq, 1\text{ M}) \parallel \text{M}^{4+}(aq), \text{M}^{2+}(aq) \mid \text{Pt}(s) the cell electromotive force is Ecell=0.092 VE_{\text{cell}} = 0.092\text{ V} when [M2+(aq)][M4+(aq)]=10x\frac{[\text{M}^{2+}(aq)]}{[\text{M}^{4+}(aq)]} = 10^x.

Given that EM4+/M2+0=0.151 VE^0_{\text{M}^{4+}/\text{M}^{2+}} = 0.151\text{ V} and 2.303RTF=0.059 V2.303 \frac{RT}{F} = 0.059\text{ V}, evaluate the value of xx.

Options

A

-2

B

-1

C

1

D

2

Correct

Topics & Concepts

ElectrochemistryElectrochemical CellNernst Equation

Step-by-Step Solution

To determine the value of the exponent xx, we apply the Nernst equation to the given electrochemical cell.

1. Identify the Half-Cell Reactions and Overall Reaction

The representation of the cell is: Pt(s)∣H2(g,1 bar)∣H+(aq,1 M)∥M4+(aq),M2+(aq)∣Pt(s)\text{Pt}(s) \mid \text{H}_2(g, 1\text{ bar}) \mid \text{H}^+(aq, 1\text{ M}) \parallel \text{M}^{4+}(aq), \text{M}^{2+}(aq) \mid \text{Pt}(s)

  • Anode (Oxidation Half-Reaction): H2(g)→2H+(aq)+2e−(EH+/H20=0.000 V)\text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \quad \left(E^0_{\text{H}^+/\text{H}_2} = 0.000\text{ V}\right)

  • Cathode (Reduction Half-Reaction): M4+(aq)+2e−→M2+(aq)(EM4+/M2+0=0.151 V)\text{M}^{4+}(aq) + 2e^- \rightarrow \text{M}^{2+}(aq) \quad \left(E^0_{\text{M}^{4+}/\text{M}^{2+}} = 0.151\text{ V}\right)

  • Overall Cell Reaction: H2(g)+M4+(aq)→2H+(aq)+M2+(aq)\text{H}_2(g) + \text{M}^{4+}(aq) \rightarrow 2\text{H}^+(aq) + \text{M}^{2+}(aq)

From the balanced equation, the number of moles of electrons transferred is n=2n = 2.

2. Calculate the Standard Cell Potential (Ecell0E^0_{\text{cell}})

Ecell0=Ecathode0−Eanode0E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} Ecell0=0.151 V−0.000 V=0.151 VE^0_{\text{cell}} = 0.151\text{ V} - 0.000\text{ V} = 0.151\text{ V}

3. Reaction Quotient (QQ)

The reaction quotient QQ for the cell reaction is given by: Q=[H+]2 [M2+]pH2 [M4+]Q = \frac{[\text{H}^+]^2 \, [\text{M}^{2+}]}{p_{\text{H}_2} \, [\text{M}^{4+}]}

Given:

  • [H+]=1 M[\text{H}^+] = 1\text{ M}
  • pH2=1 barp_{\text{H}_2} = 1\text{ bar}
  • [M2+][M4+]=10x\frac{[\text{M}^{2+}]}{[\text{M}^{4+}]} = 10^x

Substituting these values into the reaction quotient expression: Q=(1)21×10x=10xQ = \frac{(1)^2}{1} \times 10^x = 10^x

4. Apply the Nernst Equation

The Nernst equation for the cell at 298 K298\text{ K} is: Ecell=Ecell0−2.303RTnFlog⁡10QE_{\text{cell}} = E^0_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} Q

Given Ecell=0.092 VE_{\text{cell}} = 0.092\text{ V}, 2.303RTF=0.059 V2.303 \frac{RT}{F} = 0.059\text{ V}, and n=2n = 2: 0.092=0.151−0.0592log⁡10(10x)0.092 = 0.151 - \frac{0.059}{2} \log_{10}(10^x)

Since log⁡10(10x)=x\log_{10}(10^x) = x: 0.092=0.151−0.0592⋅x0.092 = 0.151 - \frac{0.059}{2} \cdot x

Rearranging the equation to solve for xx: 0.0592⋅x=0.151−0.092\frac{0.059}{2} \cdot x = 0.151 - 0.092 0.0592⋅x=0.059\frac{0.059}{2} \cdot x = 0.059 x=2x = 2

Thus, the value of xx is 2, which corresponds to Option D.

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