On passing 'C ampere of current for time 't' sec through 1 L of 2 (M) CuSO₄ solution (atomic weight of Cu= 63. 5), the amount 'm' of Cu (in gram) deposited on cathode will be

• m = Ct / (63.5 x 96500)

• m = Ct / (31.25 x 96500)

• m = (C x 96500) / (31.25xt)

• m = (31.25 X C x t) / 96500

When 96.5 C of electricity is passed through a solution of silver nitrate (at wt. of Ag = 107. 8 7 which is approx. 108), the amount of silver deposted is

• 5.8 mg

• 10.8 mg

• 15.8 mg

• 20.8 mg

The standard reduction potentials at 298 K for the following half reactions are given against each

                          $$\begin{gathered} {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Zn}\left(\mathrm{s}\right) -0.762 \mathrm{V} \\ {\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Cr}\left(\mathrm{s}\right) - 0.740 \mathrm{V} \\ 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons {\mathrm{H}}_2\left(\mathrm{s}\right) 0.000 \mathrm{V} \\ {\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \rightleftharpoons {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right) 0.770 \mathrm{V} \end{gathered}$$

Which is the strongest reducing agent ?

• Zn(s)

• Cr(s)

• H₂(s)

• Fe²⁺ (aq)

On passing 3A of electricity for 50 min, 1.8 g metal is deposited, the equivalent mass of metal is

• 9.3

• 19.3

• 38.3

• 39.9

The hydrogen electrode is dipped in a solution of pH = 3 at 25°C. The potential of the cell would be [use value of $\frac{2.303 \mathrm{RT}}{\mathrm{F}}$ = 0.59 V]

• 0.059 V

• 0.088 V

• 0.178 V

• -0.177 V

The equivalent conductivity of a solution containing 2.54 g of CuSO₄, per L is 91.0 W^-1 cm²eq^-1. Its conductivity would be

• $2.9 \times {10}^{-3}{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$

• $1.8 \times {10}^{-2} {\mathrm{\Omega }}^{-1} {\mathrm{cm}}^{-1}$

• $2.4 \times {10}^{-4} {\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$

• $3.6 \times {10}^{-3}{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$

$$\begin{gathered} {\mathrm{MnO}}_4^{-} + 8{\mathrm{H}}^{-} + 5{\mathrm{e}}^{-} \to {\mathrm{Mn}}^{2+} + 4{\mathrm{H}}_2\mathrm{O}; \mathrm{E}° = 1.51\mathrm{V} \\ {\mathrm{MnO}}_2 + 4{\mathrm{H}}^{+} + 2{\mathrm{e}}^{-} \to {\mathrm{Mn}}^{2+} + 2{\mathrm{H}}_2\mathrm{O}; \mathrm{E}° = 1.23 \mathrm{V} \mathrm{E}{°}_{{\mathrm{MnO}}_4^{-}|{\mathrm{MnO}}_2} \mathrm{is} \end{gathered}$$

• 1.70 V

• 0.91 V

• 1.37 V

• 0.548 V

388.

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2(SO₄)₃ ? Given that ${\mathrm{\Lambda }}_{{\mathrm{Al}}^{3+}}^{°}$ and ${\mathrm{\Lambda }}_{{\mathrm{SO}}_4^{2-}}^{°}$ are the equivalent conductances at infinite dilution of the respective ions?

• $2{\mathrm{\Lambda }}_{{\mathrm{Al}}^{3+}}^{°} + 3{\mathrm{\Lambda }}_{{\mathrm{SO}}_4^{2-}}^{°}$

• ${\mathrm{\Lambda }}_{{\mathrm{Al}}^{3+}}^{°} + {\mathrm{\Lambda }}_{{\mathrm{SO}}_4^{2-}}^{°}$

• $({\mathrm{\Lambda }}_{{\mathrm{Al}}^{3+}}^{°}+ 3{\mathrm{\Lambda }}_{{\mathrm{SO}}_4^{2-}}^{°} )\times 6$

• $\frac{1}{3}{\mathrm{\Lambda }}_{{\mathrm{Al}}^{3+}}^{°}+ \frac{1}{2}{\mathrm{\Lambda }}_{{\mathrm{SO}}_4^{2-}}^{°}$

The equivalent conductances of two ions at infinite dilution in water at 25°C are given below

          $$\begin{gathered} {\mathrm{\Lambda }}_{{\mathrm{Ba}}^{2+}}^{°}= 127.00 {\mathrm{Scm}}^2/\mathrm{equiv}. \\ {\mathrm{\Lambda }}_{{\mathrm{Cl}}^{-}}^{°} = 76.00 {\mathrm{Scm}}^2/ \mathrm{equiv}. \end{gathered}$$

The equivalent conductance (in Scm²/equiv) of BaCl₂ at infinite dilution will be

• 203

• 279

• 205.5

• 139.5

The EMF of the cell,

Mg | Mg²+(0.01M) || Sn²⁺(0.1M)| Sn at 298K is $({\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^{°}= -2.34 \mathrm{V}, {\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{°}= -0.14\mathrm{V})$

• 2.17 V

• 2.23V

• 2.51 V

• 2.45 V