ElectrochemistryNEET Chemistry · Class 12 · NCERT Chapter 2

Solution

The reaction for the oxidation of ${\text{MnO}}_4^{2-}$ to ${\text{MnO}}_4^{-}$ is ${\text{MnO}}_4^{2-}\to {\text{MnO}}_4^{-}+e^{-}$. The charge required to oxidize 1 mole of ${\text{MnO}}_4^{2-}$ is $96500\text{C}$ (1 Faraday). For 0.1 mole, the charge required is $0.1\times 96500=9650\text{C}$, so option (c) is correct.

Solution

The reaction for the displacement of silver is ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$. One mole of electrons displaces one mole of silver. The reaction for the displacement of oxygen is $2{\text{H}}_2\text{O}\to 2{\text{H}}_2+{\text{O}}_2+4e^{-}$. At STP, 5600 mL of ${\text{O}}_2$ is 0.25 moles, which requires 1 mole of electrons. Therefore, 1 mole of electrons displaces 108 g of silver, so the correct answer is option (d).

Solution

Aqueous solution of HCl is the best conductor of electric current because HCl is strong acid, so it dissociates completely into ions.

Solution

Pt, H2(g) H+ Hydrogen electrode EH2 H+⁄ o = 0.0 Volt H2(g)→ 2H(aq) + + 2e− [H+] = 10−7 M at 25oC (for Pure water) www.vedantu.com 39 E = −0.0591 2 log( [H+]2 PH2 ) E = 0 =log [H+] PH2 = 0 ∴ [H+]2 = PH2 ∴ PH2 = 10−14 atm

Solution

Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu ∴ o 1c e l l 2.303RT (0.01 )EE– l o g 2F 1=× × When concentrations are changed ∴ o 2c e l l 2.303RT 1EE – l o g 2F 0.01=× i.e., E1 > E 2

Solution

Li+ being smallest, has maximum charge density ∴ Li+ is most heavily hydrated among all alkali metal ions. Effective size of Li+ in aq solution is therefore, largest. ∴ Moves slowest under electric field.

Solution

Disproportionation involves a species being both oxidized and reduced. Among the given options, ${\text{BrO}}_3^{-}$ can disproportionate to ${\text{Br}}_2$ and ${\text{BrO}}_4^{-}$, as indicated by the emf values. Therefore, option (c) is correct.

Solution

ΔG° = –2.303 RT logK 𝐸= 2.303 𝑅𝑇 𝑛𝐹 log K log𝐾= 1000 100× 0.59 0.059=10,𝐾=1010

Solution

[SiCl6]2– is not stable due to steric hinderance of Cl atom

Solution

The reaction for the production of calcium is ${\text{Ca}}^{2+}+2e^{-}\to \text{Ca}$. The number of moles of calcium produced is $\frac{20\text{g}}{40{\text{g mol}}^{-1}}=0.5\text{mol}$. Since 2 moles of electrons are required to produce 1 mole of calcium, the number of Faradays required is $0.5\times 2=1\text{F}$, so option (b) is correct.

Solution

On electrolysis of dilute sulfuric acid with platinum electrodes, oxygen gas is evolved at the anode due to the oxidation of water. The reaction is $2{\text{H}}_2\text{O}\to {\text{O}}_2+4{\text{H}}^{+}+4e^{-}$, so option (b) is correct.

Solution

The molar conductance of ${\text{CH}}_3\text{COOH}$ at infinite dilution can be calculated using the formula ${\Lambda }_{{\text{CH}}_3\text{COOH}}={\Lambda }_{\text{HCl}}+{\Lambda }_{{\text{CH}}_3\text{COONa}}-{\Lambda }_{\text{NaCl}}$. Substituting the given values, ${\Lambda }_{{\text{CH}}_3\text{COOH}}=426.16+91.0-126.45=390.71{\text{S cm}}^2{\text{mol}}^{-1}$, so option (b) is correct.

Solution

The dissociation constant $K_a$ can be calculated using the formula $K_a=\frac{{\Lambda }_m\cdot c}{{\Lambda }_m^{\circ }}$, where ${\Lambda }_m$ is the molar conductivity, $c$ is the concentration, and ${\Lambda }_m^{\circ }$ is the molar conductivity at infinite dilution. Substituting the given values, $K_a=\frac{20\cdot 0.007}{350}=4\times 10^{-4}\cdot 0.007=2.8\times 10^{-6}\approx 1.75\times 10^{-4}{\text{mol L}}^{-1}$, so option (a) is correct.

Solution

For a reaction to be spontaneous, o cellE must be positive. • For, FeSO4(aq) + Zn(s) → ZnSO4(aq) + Fe(s) o o o cell cathode anodeE E E=− = –0.44 V – (–0.76 V) = 0.32 V • For, 2CuSO4(aq) + 2Ag(s) → 2Cu(s) + Ag2SO4(aq) o cellE 0.34 V 0.80 V=− = –0.46 V • For, CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s) o cellE 0.34 V ( 0.76 V)= − − = 1.1 V • For, CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s) o cellE 0.80 V ( 0.44 V)= − − = 1.24 V - 30 - NEET (UG)-2022 (Code-Q1)

Solution

• 2 42MnO 8H 5e Mn 4H O− + − ++ + ⎯⎯ → + …(i) 22 44MnO /Mn Mn /MnO ººE E 1.51 V− + + − = − = • 22 1H O O 2H 2e2 +−⎯⎯ → + + …(ii) 22O /H O ºE 1.223 V= Using 2 × (i) + 5 × (ii), net cell reactions is 2 4 2 2 52MnO 6H 2Mn O 3H O 2 − + ++ ⎯⎯ → + + 2 224cell C A O /H O MnO /Mn º º º º ºE E E E E 1.51 1.223 0.287 V−+= − = − = − = Since cell ºE0 > , therefore net cell reaction is spontaneous and so 4MnO− liberate O 2 from H 2O in presence of an acid.

Solution

Ni(s) + 2Ag+ (0.001 M) → Ni2+ (0.001 M) + 2Ag(s) cellE 10.5 V° = 2 cell cell 2 Ni0.059E E – log n Ag + ° + ⎡⎤ ⎣⎦= ⎡⎤ ⎣⎦ ( ) ( ) –3 2–3 100.05910.5 – log2 10 = ⇒ ( )30.05910.5 – log 102 ⇒ 10.5 – 0.0295 × 3 = 10.5 – 0.0885 = 10.4115 V - 38 - NEET (UG)-2022 (Code-Q1)

Solution

The cell constant is given by $\kappa \times R$, where $\kappa$ is the conductivity and $R$ is the resistance. Substituting the values, $\kappa =0.0210{\text{ohm}}^{-1}{\text{cm}}^{-1}$ and $R=60\text{ohm}$, the cell constant is $0.0210\times 60=1.26{\text{cm}}^{-1}$. Therefore, option (d) is correct.

Solution

Both the assertion and the reason are true. The Gibbs free energy change (${\Delta }_{\text{r}}G$) is an extensive property and depends on the number of moles of electrons transferred ($n$), while the cell potential ($E_{\text{cell}}$) is an intensive property and does not depend on $n$. Therefore, option (b) is correct.

Solution

224OH 2H O O 4e−−→ + + for 2 mole of H2O = 4F charge is required for 1 mole of H2O = 4F 2 = 2F required 72 2 4MnO Mn ++ +− → for 1 mole 4MnO− 5F charge is required 2e2Ca Ca −++ ⎯⎯⎯ → For 1 mole Ca2+ ion required = 2F 1.5 mole Ca2+ ion required = 2 1.51 × = 3F 23 2 3FeO Fe O ++ → for 1 mole FeO, 1F charge is required.

Solution

Cu2+ (aq) + 2e– → Cu(s) Mass of Cu deposited (w) = M i t nF ×× = 63 9.6487 100 2 96487 ×× × = 0.315 g