Chemistry MCQs for NEET — Practice Questions with Answers

${\Lambda }_m={\Lambda }_m^{\infty }-b \sqrt{c}$

260 = ${\Lambda }_m^{\infty }$ – 0.5 b…

250 = ${\Lambda }_m^{\infty }$ – b … (2)

On solving & (2), we get

${\Lambda }_m^{\infty }$ = 270 Ω^– cm² mol^–1

AgCl(s) + e $\underset{}{\overset{}{\to }}$ Ag(s) + Cl^–E° = 22 V

1/₂Cl₂ + e $\underset{}{\overset{}{\to }}$ Cl^– E° = 1.36 V

We get

Ag(s) + $\frac{1}{2}$Cl₂(g) $\underset{}{\overset{}{\to }}$ AgCl(s)E°_cell = 1.14 V

∴ ΔG = –nEF° = – (96500) (1.4) = –110 KJ/mol

if reduction potential of metal ion is greater then O₂/H₂O couple, the ion is stable in water. So Co³⁺ is stable in water

[Hint: Reverse of (2) & (3) is spontaneous; weakest Oxidizing Agent here is Mg²⁺]

∵ ΔH = ΔG + T ΔS

= –nFE_cell + nFT ${\left(\frac{{\displaystyle \partial \mathrm{E}}}{{\displaystyle \partial \mathrm{T}}}\right)}_{\mathrm{p}}$

= –199.35 KJ/mol [n = 2 for Cd²⁺ +2e^–→ Cd & putting other values]

NowΔH < O ⇒ Exothermic reaction

⇒ Heat will be released increasing the temperature.

Q = 10 × 4825 = 4825 = 48250 C

no. of mole = $\frac{48250}{96500}$ = 0.5

Ag + $\frac{1}{2}$Cu⁺⁺ $\underset{}{\overset{}{\to }}$ Ag⁺ + $\frac{1}{2}$Cu

        2.00           2.00

      2 – 0.25        2 + 0.50

E_cell –E°_cell – $\frac{0.0591}{1}$ log $\frac{[{\mathrm{Ag}}^{\mathrm{+}}\mathrm{]}}{[{\mathrm{Cu}}^{++}{\mathrm{]}}^{1/2}}$

E₁ = E°_ell – $\frac{0.0591}{1}$ log $\frac{2.00}{{(2.00)}^{1/2}}$

E₂ = E°_ell – $\frac{0.0591}{1}$log $\frac{2.50}{{(1.75)}^{1/2}}$

ΔE = E₂ –E₁ = $\frac{0.0591}{1}$ $\left[\log \hspace{0.17em}\sqrt{2}\mathrm{\hspace{0.17em}}-\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{{\displaystyle 2.50}}{{\displaystyle \sqrt{1.75}}}\right]\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{0.0591}{1}$[log 1.41 – log 1.88]

= $\frac{0.0591}{1}$ [0.1492 –0.2742] = – $\frac{0.0591}{1}$ × 0.125 = –.00738 V

Conductivity is high due to [H⁺]

Anode $\left\{\begin{array}{c}2{\mathrm{H}}_{\mathrm{2}}{\mathrm{SO}}_{\mathrm{4}}\mathrm{\hspace{0.17em}}\to \hspace{0.17em}{\mathrm{H}}_{\mathrm{2}}{\mathrm{S}}_{\mathrm{2}}{\mathrm{O}}_{\mathrm{8}}\hspace{0.17em}+\hspace{0.17em}2{\mathrm{H}}^{\mathrm{+}}\hspace{0.17em}+\hspace{0.17em}2{\mathrm{e}}^{\mathrm{-}}\\ 2{\mathrm{H}}_{\mathrm{2}}\mathrm{O}\hspace{0.17em}\to \hspace{0.17em}{\mathrm{O}}_{\mathrm{2}}\hspace{0.17em}+\hspace{0.17em}4{\mathrm{H}}^{\mathrm{+}}\hspace{0.17em}+\hspace{0.17em}4{\mathrm{e}}^{\mathrm{-}}\end{array}\right\}$

Cathode {2H₂O → H₂ + 2OH^––2e^–} × 3.

Net: 2H₂SO₄ + 8H₂O → H₂S₂O8 + O₂ + 3H₂ + 6H⁺ + 6OH^–

Hence ratio of ${}^{\mathrm{n}}{\mathrm{O}}_{\mathrm{2}}$ and ${}^{\mathrm{n}}{\mathrm{H}}_{\mathrm{2}}$ is 1: 3.

Mass of the substance deposited at the cathode 

              $m=zit$

                 =$z\left(\frac{W}{V}\right)t$

                =0.367$\times {10}^{-6}\times \frac{100\times {10}^3}{125}\times 60$

                =$17.6\times {10}^{-3}kg$

B). 2F = 1 mole Fe deposited from Fe²⁺

1 F = $\frac{1}{2}$ mol Fe    deposited from F$e^{2+}=28g$

3 F = 1 mol Fe        deposited from F$e^{3+}$

1 F = $\frac{1}{3}$ mol Fe   deposited from F$e^{3+}=18.6g$