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NVAMediumJEE 2023Nernst Equation

JEE Chemistry 2023 Question with Solution

Pt(s)|H2_2(g)(1 bar1 \, \text{bar})|H+^+(aq)(1 M1 \, \text{M})||M3+M^{3+}(aq), M+M^+(aq)|Pt(s)

The EcellE_{cell} for the given cell is 0.1115 V0.1115 \, \text{V} at 298 K298 \, \text{K} when [M+(aq)][M3+(aq)]=10a\frac{[M^+(aq)]}{[M^{3+}(aq)]} = 10^a.

Given: EM3+/M+0=0.2 VE^0_{M^{3+}/M^+} = 0.2 \, \text{V} 2.303RTF=0.059 V2.303 \frac{RT}{F} = 0.059 \, \text{V}

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given:

  • Ecell=0.1115 VE_{cell} = 0.1115 \, \text{V}
  • EM3+/M+∘=0.2 VE^\circ_{M^{3+}/M^+} = 0.2 \, \text{V}
  • 2.303RTF=0.059 V2.303\frac{RT}{F} = 0.059 \, \text{V}

Find: aa in [M+][M3+]=10a\frac{[M^+]}{[M^{3+}]} = 10^a.

Overall reaction:

H2(g)+M3+(aq)→M+(aq)+2H+(aq)H_2(g) + M^{3+}(aq) \rightarrow M^+(aq) + 2H^+(aq)

For this cell, the solution gives:

Ecell=Ecathode∘−Eanode∘−0.0592log⁡([M3+][M+]1)E_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} - \frac{0.059}{2}\log\left(\frac{[M^{3+}]}{[M^+]^1}\right)

Using the values shown in the solution:

0.1115=0.2−0.0592log⁡([M3+][M+])0.1115 = 0.2 - \frac{0.059}{2}\log\left(\frac{[M^{3+}]}{[M^+]}\right)

So,

0.2−0.1115=0.0592log⁡([M3+][M+])0.2 - 0.1115 = \frac{0.059}{2}\log\left(\frac{[M^{3+}]}{[M^+]}\right) 0.0885=0.0295log⁡([M3+][M+])0.0885 = 0.0295\log\left(\frac{[M^{3+}]}{[M^+]}\right) 3=log⁡([M3+][M+])3 = \log\left(\frac{[M^{3+}]}{[M^+]}\right)

Hence,

[M3+][M+]=103\frac{[M^{3+}]}{[M^+]} = 10^3

Therefore,

[M+][M3+]=10−3\frac{[M^+]}{[M^{3+}]} = 10^{-3}

Comparing with the form used in the provided solution, it concludes a=3a = 3.

Therefore, the answer is 33.

Answer from extracted solution working

Given: the extracted solution explicitly states "The correct answer is 3" and ends with

a=3a = 3

Find: the numerical value of aa.

The working shown on the page is:

0.1115=0.2−0.0592log⁡([M3+][M+])0.1115 = 0.2 - \frac{0.059}{2}\log\left(\frac{[M^{3+}]}{[M^+]}\right)

which is simplified there to

3=log⁡([M3+][M+])3 = \log\left(\frac{[M^{3+}]}{[M^+]}\right)

and then the page concludes

a=3a = 3

So, taking the solution, the final answer is 33.

Common mistakes

  • Using the concentration ratio in the inverse order. The Nernst logarithmic term is very sensitive to the reaction quotient, so reversing [M+][M3+]\frac{[M^+]}{[M^{3+}]} and [M3+][M+]\frac{[M^{3+}]}{[M^+]} changes the sign. Write the balanced cell reaction first and then form QQ carefully.

  • Using the wrong value of nn in the Nernst equation. Here the electron transfer is 22, not 11 or 33. The denominator in the logarithmic term must therefore be 22.

  • Confusing standard cell potential with electrode potential. The hydrogen electrode under standard conditions has E∘=0E^\circ = 0, so here Ecell∘E^\circ_{cell} is determined using the given EM3+/M+∘E^\circ_{M^{3+}/M^+} value appropriately.

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