Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electrochemistry question

2022 · 29 Jul · Shift 2 · Q18
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Electrochemistry
  5. /2022 · 29 Jul · Shift 2 · Q18

Electrochemistry question

2022 · 29 Jul · Shift 2 · Q18

JEE MainChemistryElectrochemistryNumerical+4 / −1
For a cell, Cu(s)∣Cu2+(0.001 M)∥Ag+(0.01 M)∣Ag(s)\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(0.001 \,\mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \,\mathrm{M})\right| \mathrm{Ag}(\mathrm{s})Cu(s)​Cu2+(0.001M)∥Ag+(0.01M)​Ag(s) the cell potential is found to be 0.43 V0.43 \mathrm{~V}0.43 V at 298 K298 \mathrm{~K}298 K. The magnitude of standard electrode potential for Cu2+/Cu\mathrm{Cu}^{2+} / \mathrm{Cu}Cu2+/Cu is ‾\underline{\hspace{2cm}}​×10−2 V\times 10^{-2} \mathrm{~V}×10−2 V. [Given : EAg+/AgΘE_{A{g^ + }/Ag}^\ThetaEAg+/AgΘ​= 0.80 V and 2.303RTF{{2.303RT} \over F}F2.303RT​ = 0.06 V]
Numerical answer
View written solutionFree

Correct answer: 34

  1. Write the cell reaction

Given cell: Cu(s) ∣ Cu2+(0.001 M) ∥ Ag+(0.01 M) ∣ Ag(s)\mathrm{Cu}(s)\,|\,\mathrm{Cu}^{2+}(0.001\,M)\,\|\,\mathrm{Ag}^+(0.01\,M)\,|\,\mathrm{Ag}(s)Cu(s)∣Cu2+(0.001M)∥Ag+(0.01M)∣Ag(s)

Left electrode is anode, right electrode is cathode.

So,

  • Anode: Cu(s)→Cu2++2e−\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+} + 2e^-Cu(s)→Cu2++2e−
  • Cathode: Ag++e−→Ag(s)\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}(s)Ag++e−→Ag(s)

Balancing electrons: 2Ag++Cu(s)→2Ag(s)+Cu2+2\mathrm{Ag}^+ + \mathrm{Cu}(s) \rightarrow 2\mathrm{Ag}(s) + \mathrm{Cu}^{2+}2Ag++Cu(s)→2Ag(s)+Cu2+

Hence, number of electrons transferred is n=2n=2n=2

  1. Reaction quotient

For the reaction 2Ag++Cu(s)→2Ag(s)+Cu2+2\mathrm{Ag}^+ + \mathrm{Cu}(s) \rightarrow 2\mathrm{Ag}(s) + \mathrm{Cu}^{2+}2Ag++Cu(s)→2Ag(s)+Cu2+

The reaction quotient is Q=[Cu2+][Ag+]2Q = \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2}Q=[Ag+]2[Cu2+]​

Substitute the concentrations: Q=0.001(0.01)2=10−310−4=10Q = \frac{0.001}{(0.01)^2} = \frac{10^{-3}}{10^{-4}} = 10Q=(0.01)20.001​=10−410−3​=10

  1. Apply Nernst equation for the cell

At 298 K298\,K298K, Ecell=Ecell∘−0.06nlog⁡QE_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.06}{n}\log QEcell​=Ecell∘​−n0.06​logQ

Given: Ecell=0.43 V,n=2,Q=10E_{\text{cell}} = 0.43\,V, \quad n=2, \quad Q=10Ecell​=0.43V,n=2,Q=10

Since log⁡10=1\log 10 = 1log10=1, 0.43=Ecell∘−0.062(1)0.43 = E_{\text{cell}}^\circ - \frac{0.06}{2}(1)0.43=Ecell∘​−20.06​(1) 0.43=Ecell∘−0.030.43 = E_{\text{cell}}^\circ - 0.030.43=Ecell∘​−0.03 Ecell∘=0.46 VE_{\text{cell}}^\circ = 0.46\,VEcell∘​=0.46V

  1. Relate standard cell potential to electrode potentials

Ecell∘=Ecathode∘−Eanode∘E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circEcell∘​=Ecathode∘​−Eanode∘​

Here,

  • Cathode: Ag+/Ag\mathrm{Ag}^+/\mathrm{Ag}Ag+/Ag with E∘=0.80 VE^\circ = 0.80\,VE∘=0.80V
  • Anode: Cu2+/Cu\mathrm{Cu}^{2+}/\mathrm{Cu}Cu2+/Cu with E∘=?E^\circ = ?E∘=?

So, 0.46=0.80−ECu2+/Cu∘0.46 = 0.80 - E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}0.46=0.80−ECu2+/Cu∘​

Thus, ECu2+/Cu∘=0.80−0.46=0.34 VE^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.80 - 0.46 = 0.34\,VECu2+/Cu∘​=0.80−0.46=0.34V

  1. Convert into the required form

They ask for the magnitude in the form: ‾×10−2 V\underline{\hspace{2cm}} \times 10^{-2}\,V​×10−2V

Now, 0.34 V=34×10−2 V0.34\,V = 34 \times 10^{-2}\,V0.34V=34×10−2V

So the required integer is 34\boxed{34}34​

  1. Comparison with stored answer

Derived answer = 343434

Stored correct answer = 343434

Hence, the derived answer agrees with the stored answer.

PreviousNext

More from Electrochemistry

  • A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced a STP is ​ cm3. (Nearest integer) [Given : Faraday…2022 · Numerical
  • The cell potential for the given cell at 298 K Pt| H2​ (g, 1 bar) | H+ (aq) || Cu2+ (aq) | Cu(s) is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10 − x M. The value of x is ​…2022 · Numerical
  • In which of the following half cells, electrochemical reaction is pH dependent?2022 · MCQ
  • If the conductivity of mercury at 0 ∘ C is 1.07 × 106 S m − 1 and the resistance of a cell containing mercury is 0.243 Ω, then the cell constant of the cell is x × 104 m − 1. The value of x is ​…2021 · Numerical
  • A 5.0 m mol dm − 3 aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm − 1. The molar conductivity of this solution is ​ mSm2 mol − 1. (Round off to the…2021 · Numerical
  • A KCl solution of conductivity 0.14 S m − 1 shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω. The conductivity of the HCl solution is ​×…2021 · Numerical
  • For the reaction 2Fe3+(aq) + 2I −(aq) → 2Fe2+(aq) + I2​(s) the magnitude of the standard molar Gibbs free energy change, Δ rG mo​=−​ kJ (Round off to the Nearest Integer). [EFe2+/Fe(s)o​=−0.440V;EI2​/2I−o​=0.539V;​EFe3+/Fe(s)o​=−0.036VF=96500C​]…2021 · Numerical
  • The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm2 mol − 1 respectively. The molar conductivity at infinite dilution of barium sulphate is ​…2021 · Numerical