Given the following molar conductivities at ${25}^{°}C:,<mspace\; width="1em"></mspace>HCl<mspace\; width="1em"></mspace>426<mspace\; width="1em"></mspace>{\Omega }^{-1}<mspace\; width="1em"></mspace>c{m}^2<mspace\; width="1em"></mspace>mo{l}^{-1};$
$$$NaCl<mspace\; width="1em"></mspace>126<mspace\; width="1em"></mspace>{\Omega }^{-1}<mspace\; width="1em"></mspace>c{m}^2<mspace\; width="1em"></mspace>mo{l}^{-1};<mspace\; width="1em"></mspace>NaC<mspace\; width="1em"></mspace>(sodium<mspace\; width="1em"></mspace>coronate)$$83<mspace\; width="1em"></mspace>{\Omega }^{-1}<mspace\; width="1em"></mspace>c{m}^2<mspace\; width="1em"></mspace>mo{l}^{-1}$.  what is the ionization constant of crotonic acid? If the conductivity of a 0.001 M crotonic acid solution is 3.83$\times$${10}^{-5}<mspace\; width="1em"></mspace>{\Omega }^{-1}c{m}^{-1}$?

(1) ${10}^{-5}$                 (2) $1.11\times {10}^{-5}$

(3) $1.11\times {10}^{-4}$      (4)0.01

Which of the following solutions has the highest equivalent conductance?

1. 0.01 M NaCl

2. 0.05 M NaCl

3. 0.005 M NaCl

4. 0.02 M NaCl

For a reaction,  A(s) + 2B⁺ $\to$ A²⁺ + 2B(s) ; K_C  has been found to be 10¹². The  ${\mathrm{E}}_{\mathrm{cell}}^{°}$ is
1. 0.354 V

2. 0.708V

3. 0.0098 V

4. 1.36V

A fuel cell develops an electrical potential from the coombustion of butane at 1 bar and 298 K

$C_4{H}_{10}\left(g\right)+6.5O_2\left(g\right)\to 4C{O}_2\left(g\right)+5H_{2}O\left(l\right);$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>{∆}_r{G}^{°}=-2746<mspace\; width="1em"></mspace>kJ/mol$

What is $E^{°}$ of a cell?

(1) 4.74 V

(2) 0.547 V

(3) 4.37 V

(4) 1.09 V

The electrolysis of acetate solution produces ethane according to reaction:

$2C{H}_{3}CO{O}^{-}\to C_2{H}_6\left(g\right)+2C{O}_2\left(g\right)+2e^{-}$

The current efficiency of the process is 80%.  What volume of gases would be produced at 27$°$C and 740 torr, if the current of 0.5 amp is passed through the solution for 96.45 min?

(1) 6.0 L                   (2) 0.60 L                   (3) 1.365 L                       (4) 0.91 L

How many second would it take to produce enough aluminum by the Hall process to make a case of 24 cans of aluminum soft-drink, if each can uses 5.0 g of Al, a current of 9650 amp is employed, and the current efficiency of the cell is 90.0%:

(1) 203.2

(2) 148.14

(3) 333

(4) 6.17

An excess of liquid Hg was added to 10^–3 M acidified solution of ${\mathrm{Fe}}^{3+}$ ions. It was found that only 5% of the ions remained as Fe⁺³ at equilibrium at 25ºC. What is Eº for 2Hg/Hg₂⁺² at 25ºC for-                     

   2Hg + 2Fe⁺³   $\rightleftharpoons$   ${\mathrm{Hg}}_2^{2+}$ + $2{\mathrm{Fe}}^{+2}$

and   ${\mathrm{E}}_{{\mathrm{Fe}}^{+2}/{\mathrm{Fe}}^{+3}}^0$   = 0.77 V

1. – 1.347 V               

2. – 0.793 V

3. – 0.125 V               

4. – 1.110V

The reduction potential of the two half cell
reactions (occuring in an electrochemical cell)
are
PbSO₄(s) + 2e^– $\to$ Pb(s) + ${\mathrm{SO}}_4^{2-}$(aq) (E°= –0.31V)
Ag⁺(aq) + e^– $\to$ Ag(s) (E° = 0.80V)

The feasible reaction will be
1. Pb(s) + ${\mathrm{SO}}_4^{2-}$(aq) + 2 Ag⁺(aq)$\to$  2Ag (s) + PbSO₄ (s)
2. PbSO₄(s) + Ag⁺ (aq) $\to$ Pb(s)+${\mathrm{SO}}_4^{2-}$(aq) + 2Ag(s)
3. Pb(s) + ${\mathrm{SO}}_4^{2-}$ (s) + Ag(s) $\to$Ag⁺(aq) + PbSO₄(s)
4. PbSO₄(s) +Ag(s) $\to$ Ag⁺(aq) + Pb(s) +${\mathrm{SO}}_4^{2-}$ (aq)

Adding powdered lead and iron to a solution that is 1.0 molar each in ${\mathrm{Pb}}^{2+}$ and ${\mathrm{Fe}}^{2+}$ ions, would result to a reaction in which

     $\begin{array}{rl}& {\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\circ }=-0.44\mathrm{V}\\ & {\mathrm{E}}_{{\mathrm{Pb}}^{2+}/\mathrm{Pb}}^0=-0.13\mathrm{V}\end{array}$

1.  Conc. of both ${\mathrm{Pb}}^{2+}$ and ${\mathrm{Fe}}^{2+}$ are increased

2.  More lead and iron are formed

3.  More of iron and ${\mathrm{Pb}}^{2+}$ ions are formed

4.  More of lead and ${\mathrm{Fe}}^{2+}$ ions are formed

$$
For the Cell \(Pt(s)|| Br^-(aq)(0.010M)| Br_2(l)|| H^+(aq) (0.0030M) |H_2(g) (1 bar) | Pt(s)\)

If the concentration of ${\mathrm{Br}}^{-}$ becomes 2 times and the concentration of ${\mathrm{H}}^{+}$ becomes half of the initial value, then emf of the cell

1.  Doubles

2.  Four times

3.  Eight times

4.  Remains the same