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NVAEasyJEE 2023Nernst Equation

JEE Chemistry 2023 Question with Solution

Consider the cell Pt(s)∣H2(g,1 atm)∣H+(aq,1 M)∣∣Fe3+(aq),Fe2+(aq)∣∣Pt(s).\text{Pt}(s) \mid \text{H}_2 (g, 1 \, \text{atm}) \mid \text{H}^+ (aq, 1\,\text{M}) \mid\mid \text{Fe}^{3+} (aq) , \text{Fe}^{2+} (aq) \mid\mid \text{Pt}(s). When the potential of the cell is 0.712 V0.712 \, \text{V} at 298 K298 \, \text{K}, the ratio [Fe2+][Fe3+]\frac{[Fe^{2+}]}{[Fe^{3+}]} is _____. (Nearest integer)

Given: Fe3++e−→Fe2+,EFe3+∣Fe2+∘=0.771 V.\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}, \quad E^\circ_{Fe^{3+}|Fe^{2+}} = 0.771 \, V. 2.303RTF=0.06 V.\frac{2.303RT}{F} = 0.06 \, V.

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: The cell is

Pt(s)∣H2(g,1 atm)∣H+(aq,1 M)∣∣Fe3+(aq),Fe2+(aq)∣∣Pt(s)\text{Pt}(s)|\text{H}_2(g, 1\,\text{atm})|\text{H}^+(aq, 1\,\text{M})||\text{Fe}^{3+}(aq), \text{Fe}^{2+}(aq)||\text{Pt}(s)

The measured cell potential is 0.712 V0.712 \, \text{V} at 298 K298 \, \text{K}.

Find: The ratio [Fe2+][Fe3+]\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.

At the anode:

H2→2H++2e−\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-

At the cathode:

Fe(aq)3++e−→Fe(aq)2+\text{Fe}^{3+}_{(aq)} + e^- \rightarrow \text{Fe}^{2+}_{(aq)}

The standard cell potential is:

E∘=EH2/H+∘+EFe3+/Fe2+∘E^\circ = E^\circ_{\text{H}_2/\text{H}^+} + E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}

Substituting the given values:

E∘=0+0.771=0.771 VE^\circ = 0 + 0.771 = 0.771\,\text{V}

Using the Nernst equation as given in the solution:

E=E∘−0.061log⁡[Fe2+][Fe3+]E = E^\circ - \frac{0.06}{1} \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}

Substitute E=0.712 VE = 0.712 \, \text{V} and E∘=0.771 VE^\circ = 0.771 \, \text{V}:

0.712=0.771−0.06log⁡[Fe2+][Fe3+]0.712 = 0.771 - 0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}

Rearranging:

0.712−0.771=−0.06log⁡[Fe2+][Fe3+]0.712 - 0.771 = -0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} −0.059=−0.06log⁡[Fe2+][Fe3+]-0.059 = -0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}

So,

log⁡[Fe2+][Fe3+]=−0.059−0.06=1\log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = \frac{-0.059}{-0.06} = 1

Therefore,

[Fe2+][Fe3+]=101=10\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 10^1 = 10

The ratio [Fe2+][Fe3+]\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} is 1010.

Step-by-step Logarithmic Rearrangement

Given: E=0.712 VE = 0.712 \, \text{V} and E∘=0.771 VE^\circ = 0.771 \, \text{V}.

Find: The value of [Fe2+][Fe3+]\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.

Start with:

E=E∘−0.06log⁡[Fe2+][Fe3+]E = E^\circ - 0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}

Substitute the values:

0.712=0.771−0.06log⁡[Fe2+][Fe3+]0.712 = 0.771 - 0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}

Move 0.7710.771 to the left side:

0.712−0.771=−0.06log⁡[Fe2+][Fe3+]0.712 - 0.771 = -0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} −0.059=−0.06log⁡[Fe2+][Fe3+]-0.059 = -0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}

Divide by −0.06-0.06:

log⁡[Fe2+][Fe3+]≈1\log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} \approx 1

Take antilog:

[Fe2+][Fe3+]=10\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 10

Hence, the required nearest integer is 1010.

Common mistakes

  • Using the wrong reaction quotient by writing [Fe3+][Fe2+]\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]} instead of [Fe2+][Fe3+]\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}. This reverses the logarithmic term and changes the final value. Use the quotient exactly as written in the Nernst expression for the reduction half-cell.

  • Substituting the hydrogen electrode potential as non-zero under the given conditions. Since the electrode is at 1 atm1 \, \text{atm} hydrogen and 1 M1 \, \text{M} acid concentration, it is the standard hydrogen electrode, so its standard potential is 00.

  • Making a sign error while rearranging 0.712=0.771−0.06log⁡[Fe2+][Fe3+]0.712 = 0.771 - 0.06 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}. If the negative signs are handled incorrectly, the logarithm becomes negative and gives the inverse ratio. Keep track of signs carefully before taking antilogarithm.

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