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NVAMediumJEE 2026Nernst Equation

JEE Chemistry 2026 Question with Solution

Consider the following electrochemical cell at 298 K298\,K:

Pt ∣ HSnO2−(aq) ∣ Sn(OH)62−(aq) ∣ OH−(aq) ∣ Bi2O3(s) ∣ Bi(s)Pt \, | \, \mathrm{HSnO_2^- (aq)} \, | \, \mathrm{Sn(OH)_6^{2-} (aq)} \, | \, \mathrm{OH^- (aq)} \, | \, \mathrm{Bi_2O_3 (s)} \, | \, \mathrm{Bi (s)}

If the reaction quotient at a given time is 10610^6, then the cell EMF (EcellE_{cell}) is _____ ×10−1\times 10^{-1} V (Nearest integer).

Given: EBi2O3/Bi,OH−∘=−0.44 V,ESn(OH)62−/HSnO2−,OH−∘=−0.90 VE^\circ_{\mathrm{Bi_2O_3/Bi,OH^-}} = -0.44\ V, \quad E^\circ_{\mathrm{Sn(OH)_6^{2-}/HSnO_2^-,OH^-}} = -0.90\ V

Answer

Correct answer:28

Step-by-step solution

Standard Method

Given: The electrochemical cell is

Pt ∣ HSnO2−(aq) ∣ Sn(OH)62−(aq) ∣ OH−(aq) ∣ Bi2O3(s) ∣ Bi(s)Pt \, | \, \mathrm{HSnO_2^- (aq)} \, | \, \mathrm{Sn(OH)_6^{2-} (aq)} \, | \, \mathrm{OH^- (aq)} \, | \, \mathrm{Bi_2O_3 (s)} \, | \, \mathrm{Bi (s)}

with Q=106Q = 10^6 at 298 K298\,\text{K}.

Standard reduction potentials are

EBi2O3/Bi,OH−∘=−0.44 VE^\circ_{\mathrm{Bi_2O_3/Bi,OH^-}} = -0.44\,\text{V}

and

ESn(OH)62−/HSnO2−,OH−∘=−0.90 VE^\circ_{\mathrm{Sn(OH)_6^{2-}/HSnO_2^-,OH^-}} = -0.90\,\text{V}

Find: The nearest integer in Ecell=_____×10−1 VE_{\text{cell}} = \_\_\_\_\_ \times 10^{-1}\,\text{V}.

Using the standard cell potential relation,

Ecell∘=Ecathode∘−Eanode∘=−0.44−(−0.90)=0.46 VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.44 - (-0.90) = 0.46\,\text{V}

The balanced overall redox reaction involves 22 electrons, so n=2n = 2.

Apply the Nernst equation:

Ecell=Ecell∘−RTnFln⁡QE_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q

Substituting the given values,

Ecell=0.46−8.314×2982×96485ln⁡(106)E_{\text{cell}} = 0.46 - \frac{8.314 \times 298}{2 \times 96485} \ln(10^6)

Using

ln⁡(106)≈13.815\ln(10^6) \approx 13.815

we get

Ecell=0.46−0.02572×13.815E_{\text{cell}} = 0.46 - \frac{0.0257}{2} \times 13.815

Therefore,

Ecell=0.46−0.1775≈0.2825 VE_{\text{cell}} = 0.46 - 0.1775 \approx 0.2825\,\text{V}

So,

Ecell≈28×10−2 V=2.8×10−1 VE_{\text{cell}} \approx 28 \times 10^{-2}\,\text{V} = 2.8 \times 10^{-1}\,\text{V}

Hence the required nearest integer is 2828 as concluded by the solution.

Using the 0.0591 form at 298 K

Given: Q=106Q = 10^6, T=298 KT = 298\,\text{K}, n=2n = 2, and

Ecell∘=−0.44−(−0.90)=0.46 VE^\circ_{\text{cell}} = -0.44 - (-0.90) = 0.46\,\text{V}

Find: The numerical value asked in the blank.

At 298 K298\,\text{K}, the Nernst equation may be written as

Ecell=Ecell∘−0.0591nlog⁡QE_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q

Now,

log⁡(106)=6\log(10^6) = 6

so

Ecell=0.46−0.05912×6E_{\text{cell}} = 0.46 - \frac{0.0591}{2} \times 6

Thus,

Ecell=0.46−0.1773≈0.2827 VE_{\text{cell}} = 0.46 - 0.1773 \approx 0.2827\,\text{V}

This is approximately

2.8×10−1 V2.8 \times 10^{-1}\,\text{V}

Therefore, the nearest integer asked for in the blank is 2828.

Common mistakes

  • Taking the wrong value of nn is the most common error. The Nernst equation requires the number of electrons transferred in the balanced overall reaction, not the oxidation-state change of only one species. Here, use n=2n = 2.

  • Using Ecell∘=Eanode∘−Ecathode∘E^\circ_{\text{cell}} = E^\circ_{\text{anode}} - E^\circ_{\text{cathode}} is incorrect. Standard cell potential is calculated as Ecathode∘−Eanode∘E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}, which gives 0.46 V0.46\,\text{V} here.

  • Mixing up ln⁡Q\ln Q and log⁡Q\log Q leads to an incorrect numerical factor. If you use natural logarithm, use RTnFln⁡Q\frac{RT}{nF} \ln Q. If you use common logarithm at 298 K298\,\text{K}, use 0.0591nlog⁡Q\frac{0.0591}{n} \log Q.

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