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Electrochemistry question

2024 · 27 Jan · Shift 1 · Q21
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  5. /2024 · 27 Jan · Shift 1 · Q21

Electrochemistry question

2024 · 27 Jan · Shift 1 · Q21

JEE MainChemistryElectrochemistryNumerical+4 / −1
The mass of silver (Molar mass of Ag:108 gmol−1\mathrm{Ag}: 108 \mathrm{~gmol}^{-1}Ag:108 gmol−1) displaced by a quantity of electricity which displaces 5600 mL5600 \mathrm{~mL}5600 mL of O2\mathrm{O}_2O2​ at S.T.P. will be ‾\underline{\hspace{2cm}}​ g.
Numerical answer
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Correct answer: 108

  1. Find moles of O2\mathrm{O_2}O2​ displaced at STP

At STP, 111 mole of gas occupies 22.4 L=22400 mL22.4\,\mathrm{L} = 22400\,\mathrm{mL}22.4L=22400mL.

Given volume of oxygen: 5600 mL=5.6 L5600\,\mathrm{mL} = 5.6\,\mathrm{L}5600mL=5.6L

So, moles of O2\mathrm{O_2}O2​ are: n(O2)=560022400=14 moln(\mathrm{O_2}) = \frac{5600}{22400} = \frac{1}{4}\,\mathrm{mol}n(O2​)=224005600​=41​mol

  1. Relate oxygen evolved to electrons transferred

At the anode, oxygen is liberated as: 2H2O→O2+4H++4e−2\mathrm{H_2O} \rightarrow \mathrm{O_2} + 4\mathrm{H^+} + 4e^-2H2​O→O2​+4H++4e−

Thus, 111 mole of O2\mathrm{O_2}O2​ corresponds to 444 moles of electrons.

Therefore, 14\frac{1}{4}41​ mole of O2\mathrm{O_2}O2​ corresponds to: 14×4=1 mol e−\frac{1}{4} \times 4 = 1\,\text{mol } e^-41​×4=1mol e−

So the quantity of electricity passed is 1 Faraday.

  1. Use this charge to find silver deposited

Silver is deposited according to: Ag++e−→Ag\mathrm{Ag^+} + e^- \rightarrow \mathrm{Ag}Ag++e−→Ag

Here, 111 mole of electrons deposits 111 mole of silver.

Since 111 mole of electrons passed, moles of silver deposited: n(Ag)=1 moln(\mathrm{Ag}) = 1\,\mathrm{mol}n(Ag)=1mol

Mass of silver: m=nM=1×108=108 gm = nM = 1 \times 108 = 108\,\mathrm{g}m=nM=1×108=108g

  1. Final Answer

108\boxed{108}108​

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