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Electrochemistry question

2023 · 30 Jan · Shift 2 · Q20
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Electrochemistry question

2023 · 30 Jan · Shift 2 · Q20

JEE MainChemistryElectrochemistryNumerical+4 / −1
The electrode potential of the following half cell at 298 KX∣X2+(0.001M)∥Y2+(0.01M)∣Y298 \mathrm{~K}\mathrm{X}\left|\mathrm{X}^{2+}(0.001 \mathrm{M}) \| \mathrm{Y}^{2+}(0.01 \mathrm{M})\right| \mathrm{Y}298 KX​X2+(0.001M)∥Y2+(0.01M)​Y is ‾\underline{\hspace{2cm}}​×10−2 V\times 10^{-2} \mathrm{~V}×10−2 V(Nearest integer) Given: EX2+∣X0=−2.36 VEY2+∣Y0=+0.36 V2.303RTF=0.06 V\mathrm{E}^{0} _ {\mathrm{X}^{2+} \mid \mathrm{X}}=-2.36 \mathrm{~V}\mathrm{E}_{\mathrm{Y}^{2+} \mid \mathrm{Y}}^{0}=+0.36 \mathrm{~V}\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}EX2+∣X0​=−2.36 VEY2+∣Y0​=+0.36 VF2.303RT​=0.06 V
Numerical answer
View written solutionFree

Correct answer: 275

  1. Identify anode and cathode

The cell is

X ∣ X2+(0.001 M) ∥ Y2+(0.01 M) ∣ Y\mathrm{X}\,|\,\mathrm{X}^{2+}(0.001\,\mathrm{M})\,\|\,\mathrm{Y}^{2+}(0.01\,\mathrm{M})\,|\,\mathrm{Y}X∣X2+(0.001M)∥Y2+(0.01M)∣Y

Given standard reduction potentials:

EX2+/X∘=−2.36 V,EY2+/Y∘=+0.36 VE^\circ_{\mathrm{X}^{2+}/\mathrm{X}}=-2.36\,\mathrm{V},\qquad E^\circ_{\mathrm{Y}^{2+}/\mathrm{Y}}=+0.36\,\mathrm{V}EX2+/X∘​=−2.36V,EY2+/Y∘​=+0.36V

Since Y2+/Y\mathrm{Y}^{2+}/\mathrm{Y}Y2+/Y has higher reduction potential, Y is cathode and X is anode.

So the overall reaction is:

X+Y2+→X2++Y\mathrm{X}+\mathrm{Y}^{2+} \rightarrow \mathrm{X}^{2+}+\mathrm{Y}X+Y2+→X2++Y

with n=2n=2n=2 electrons.

  1. Calculate standard cell potential
Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}Ecell∘​=Ecathode∘​−Eanode∘​ Ecell∘=0.36−(−2.36)=2.72 VE^\circ_{\text{cell}}=0.36-(-2.36)=2.72\,\mathrm{V}Ecell∘​=0.36−(−2.36)=2.72V
  1. Write reaction quotient

For the reaction

X+Y2+→X2++Y\mathrm{X}+\mathrm{Y}^{2+} \rightarrow \mathrm{X}^{2+}+\mathrm{Y}X+Y2+→X2++Y

Solids are omitted, hence

Q=[X2+][Y2+]=0.0010.01=0.1Q=\frac{[\mathrm{X}^{2+}]}{[\mathrm{Y}^{2+}]}=\frac{0.001}{0.01}=0.1Q=[Y2+][X2+]​=0.010.001​=0.1
  1. Apply Nernst equation

At 298 K298\,\mathrm{K}298K,

Ecell=Ecell∘−0.06nlog⁡QE_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.06}{n}\log QEcell​=Ecell∘​−n0.06​logQ

Since n=2n=2n=2,

Ecell=2.72−0.062log⁡(0.1)E_{\text{cell}}=2.72-\frac{0.06}{2}\log(0.1)Ecell​=2.72−20.06​log(0.1)

Now,

log⁡(0.1)=−1\log(0.1)=-1log(0.1)=−1

Therefore,

Ecell=2.72−0.03(−1)=2.72+0.03=2.75 VE_{\text{cell}}=2.72-0.03(-1)=2.72+0.03=2.75\,\mathrm{V}Ecell​=2.72−0.03(−1)=2.72+0.03=2.75V
  1. Express in the required form

The question asks for

‾×10−2 V\underline{\hspace{1cm}}\times 10^{-2}\,\mathrm{V}​×10−2V

Since

2.75 V=275×10−2 V2.75\,\mathrm{V}=275\times 10^{-2}\,\mathrm{V}2.75V=275×10−2V

So the required integer is:

275\boxed{275}275​
  1. Comparison with stored answer

Derived answer = 275275275

Stored correct answer = 275275275

Hence, they agree.

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