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

2023 · 29 Jan · Shift 1 · Q11
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. /2023 · 29 Jan · Shift 1 · Q11

Electrochemistry question

2023 · 29 Jan · Shift 1 · Q11

JEE MainChemistryElectrochemistryMCQ+4 / −1
The standard electrode potential (M3+/M2+)\mathrm{(M^{3+}/M^{2+})}(M3+/M2+) for V, Cr, Mn & Co are −-− 0.26 V, −-− 0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate H2\mathrm{H_2}H2​ from a dilute acid are :
  1. A
    Mn2+\mathrm{Mn^{2+}}Mn2+ and Co2+\mathrm{Co^{2+}}Co2+
  2. B
    V2+\mathrm{V^{2+}}V2+ and Mn2+\mathrm{Mn^{2+}}Mn2+
  3. C
    V2+\mathrm{V^{2+}}V2+ and Cr2+\mathrm{Cr^{2+}}Cr2+
  4. D
    Cr2+\mathrm{Cr^{2+}}Cr2+ and Co2+\mathrm{Co^{2+}}Co2+
View written solutionFree

Correct answer: C

  1. Relevant reaction for liberation of hydrogen

A metal ion in lower oxidation state, say M2+\mathrm{M^{2+}}M2+, can liberate hydrogen from dilute acid if it can reduce H+\mathrm{H^+}H+ to H2\mathrm{H_2}H2​.

The reaction is:

2M2++2H+→2M3++H22\mathrm{M^{2+}} + 2\mathrm{H^+} \rightarrow 2\mathrm{M^{3+}} + \mathrm{H_2}2M2++2H+→2M3++H2​

Here:

  • M2+→M3+\mathrm{M^{2+} \to M^{3+}}M2+→M3+ is oxidation
  • 2H++2e−→H2\mathrm{2H^+ + 2e^- \to H_2}2H++2e−→H2​ has standard reduction potential E∘=0.00 VE^\circ = 0.00\,\text{V}E∘=0.00V
  1. Condition for spontaneity

Given data are standard reduction potentials for:

M3++e−→M2+\mathrm{M^{3+} + e^- \to M^{2+}}M3++e−→M2+

If M2+\mathrm{M^{2+}}M2+ is to act as a reducing agent, it must undergo the reverse reaction:

M2+→M3++e−\mathrm{M^{2+} \to M^{3+} + e^-}M2+→M3++e−

So for the cell:

  • Cathode: 2H++2e−→H2\mathrm{2H^+ + 2e^- \to H_2}2H++2e−→H2​, E∘=0E^\circ = 0E∘=0
  • Anode: M2+→M3++e−\mathrm{M^{2+} \to M^{3+} + e^-}M2+→M3++e−

Thus,

Ecell∘=Ecathode∘−Eanode (as reduction)∘=0−E∘(M3+/M2+)E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode (as reduction)}} = 0 - E^\circ(\mathrm{M^{3+}/M^{2+}})Ecell∘​=Ecathode∘​−Eanode (as reduction)∘​=0−E∘(M3+/M2+)

For hydrogen liberation,

Ecell∘>0Rightarrow0−E∘(M3+/M2+)>0RightarrowE∘(M3+/M2+)<0E^\circ_{\text{cell}} > 0 Rightarrow 0 - E^\circ(\mathrm{M^{3+}/M^{2+}}) > 0 Rightarrow E^\circ(\mathrm{M^{3+}/M^{2+}}) < 0Ecell∘​>0Rightarrow0−E∘(M3+/M2+)>0RightarrowE∘(M3+/M2+)<0

So only those ions with negative E∘(M3+/M2+)E^\circ(\mathrm{M^{3+}/M^{2+}})E∘(M3+/M2+) can liberate H2\mathrm{H_2}H2​ from dilute acid.

  1. Check each metal ion

Given:

  • For V: E∘=−0.26 VE^\circ = -0.26\,\text{V}E∘=−0.26V
  • For Cr: E∘=−0.41 VE^\circ = -0.41\,\text{V}E∘=−0.41V
  • For Mn: E∘=+1.57 VE^\circ = +1.57\,\text{V}E∘=+1.57V
  • For Co: E∘=+1.97 VE^\circ = +1.97\,\text{V}E∘=+1.97V

Negative values are for:

  • V3+/V2+\mathrm{V^{3+}/V^{2+}}V3+/V2+
  • Cr3+/Cr2+\mathrm{Cr^{3+}/Cr^{2+}}Cr3+/Cr2+

Hence the ions that can reduce H+\mathrm{H^+}H+ to H2\mathrm{H_2}H2​ are:

V2+ and Cr2+\mathrm{V^{2+}} \text{ and } \mathrm{Cr^{2+}}V2+ and Cr2+
  1. Option-wise check

  • A: Mn2+\mathrm{Mn^{2+}}Mn2+ and Co2+\mathrm{Co^{2+}}Co2+
    Both have positive potentials, so cannot liberate H2\mathrm{H_2}H2​.

  • B: V2+\mathrm{V^{2+}}V2+ and Mn2+\mathrm{Mn^{2+}}Mn2+
    V2+\mathrm{V^{2+}}V2+ can, but Mn2+\mathrm{Mn^{2+}}Mn2+ cannot. So incorrect.

  • C: V2+\mathrm{V^{2+}}V2+ and Cr2+\mathrm{Cr^{2+}}Cr2+
    Both have negative potentials, so correct.

  • D: Cr2+\mathrm{Cr^{2+}}Cr2+ and Co2+\mathrm{Co^{2+}}Co2+
    Cr2+\mathrm{Cr^{2+}}Cr2+ can, but Co2+\mathrm{Co^{2+}}Co2+ cannot. So incorrect.

  1. Final answer

The metal ions which can liberate H2\mathrm{H_2}H2​ from dilute acid are:

V2+ and Cr2+\boxed{\mathrm{V^{2+}} \text{ and } \mathrm{Cr^{2+}}}V2+ and Cr2+​

So the correct option is C.

PreviousNext

More from Electrochemistry

  • Following figure shows dependence of molar conductance of two electrolytes on concentration. Λmo is the limiting molar conductivity. The number of incorrect​ statement(s) from the following… Includes diagram2023 · Numerical
  • The equilibrium constant for the reaction Zn(s)+Sn2+(aq) ⇌ Zn2+(aq)+Sn(s) is 1×1020 at 298 K. The magnitude of standard electrode potential of Sn/Sn2+ if EZn2+/ZnΘ​=−0.76 V…2023 · Numerical
  • Consider the cell Pt(s)​∣H2​( g,1 atm)∣H+(aq,1M)∣∣Fe3+(aq),Fe2+(aq)∣Pt(s)…2023 · Numerical
  • The electrode potential of the following half cell at 298 KX​X2+(0.001M)∥Y2+(0.01M)​Y is ​$\times 10^{-2}…2023 · Numerical
  • Which one of the following statements is correct for electrolysis of brine solution?2023 · MCQ
  • The logarithm of equilibrium constant for the reaction Pd2++4Cl−⇌PdCl42−​ is ​ (Nearest integer) Given :  F2.303R T​=0.06 VPd(aq)2+​+2e−⇌Pd(s)E⊖=0.83 V​PdCl42−​(aq)+2e−⇌Pd(s)+4Cl−(aq)E⊖=0.65 V​…2023 · Numerical
  • The resistivity of a 0.8M solution of an electrolyte is 5×10−3Ω cm. Its molar conductivity is ​×104 Ω−1 cm2 mol−1. (Nearest…2023 · Numerical
  • The cell potential for the following cell Pt |H2​(g)|H+ (aq)|| Cu2+ (0.01 M)|Cu(s) is 0.576 V at 298 K. The pH of the solution is ​. (Nearest integer) (Given : ECu2+/Cuo​=0.34 V and F2.303RT​=0.06…2022 · Numerical