In the following electrochemical cell, $M|M^{+} || X^{-}|X,$ the standard reduction potentials are $E{°}_{M+/M}= 0.52 V$ and $E{°}_{X/X^{-}} = 0.42 V at 25°C.$ Which one of the following statements is correct ?

$\left(1\right) M + X \to M^{+} + X^{-} is spon\tan eous$              

$\left(2\right) M^{+} + X^{-} \to M + X is spon\tan eous$

$\left(3\right) E_{cell} = 0.77 V$                                              

$\left(4\right) E_{cell}= – 0.77 V$

1If a 100 mL solution of 0.1M HBr is titrated using a very concentrated solution of NaOH, when the conductivity (specific conductance) of the solution at the equivalence point will 3e (assume volume change is negligible due to addition of NaOH). Calculate your answer in terms of ${10}^{-1}S{m}^{-1},$

$\left[Given {{\wedge }^{°}}_{\left(N{a}^{+}\right)}=8\times {10}^{-3}S{m}^{2}mo{l}^{-1} {{\wedge }^{°}}_{\left(B{r}^{-}\right)}=4\times {10}^{-3}S m^2 mo{l}^{-1}\right]$

(1) 6                               

(2) 12

(3) 15                             

(4) 24

The volume of gases liberated at STP when a charge of 2F is passed through aqueous solution of sodium phosphate, is :

1. 11.2L                                           

2. 44.8L

3. 33.6L                                           

4. 22.4L

It is desired to convert the energy derived from the combustion of propane into electrical energy, via a fuel cell. Given that the standard free energies of formation of $C_3{H}_8 \left(g\right), H_{2}O\left(l\right) and C{O}_2 \left(g\right)$ are -23.5, -237.2 and -394.4 $kJ mo{l}^{-1}$ respectively, calculate the standard EMF of the propane fuel cell.

(1) 1.10 V                                           

(2) 2.10V

(3) 3.12 V                                           

(4) 4.12 V

The temperature coefficient of a standard Cd-cell is $–5.0 \times {10}^{-5} V{k}^{-1}$ whose emf at $25°C$ is 1.018 V. During the cell operation, the temperature will -

(1) increase                                      

(2) decreases

(3) either                                         

(4) remains constant

Na-amalgam is prepared by electrolysis of NaCl solution using liquid Hg as a cathode. How

long should the current of 10 amp. is passed to produce 10% Na – Hg on a cathode of 10

gm Hg. (atomic mass of Na = 23).

(A) 7.77 min.                                               

(B) 9.44 min.

(C) 5.24 min.                                               

(D) 11.39 min.

The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in ${10}^6 L of a 0.1 M AgN{O}_3$ solution is 

(1) 0.5 mmol                                              

(2) 1.0 mmol

(2) 2.0 mmol                                             

(4) 2.5 mmol

The $k=$ $4.95$ $\times$ ${10}^{-5}S$ $c{m}^{-1}$ for a 0.00099 M solution. The reciprocal of the degree of dissociation of acetic acid, if ${\wedge }_m^0$ for acetic acid is 400 S $c{m}^{2}mo{l}^{-1}$ will be:

Pure water is saturated with pure solid AgCl, a silver electrode is placed in the solution and the potential is measured against normal calomel electrode at $25°C$. This experiment is then repeated with a saturated solution of AgI. If the difference in potential in the two cases is 0.177 V. What is the ratio of solubilities of AgCl and AgI at the temperature of the experiment ?

$\left(1\right) {10}^3$

$\left(2\right) {10}^6$

$\left(3\right) {10}^4$

$\left(4\right) {10}^2$

A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the ratio a : b and another hydrogen electrode Y was placed in a buffer solution of sodium acetate and acetic acid in the ratio b : a. If reduction potential values for two cells are found to be $E_{1}and E_2$ respectively with respect to standard hydrogen electrode, the $p{K}_a$ value of the acid can be given as

(1) $\frac{E_1-E_2}{0.118}$                                           

(2) $\frac{E_1+E_2}{0.118}$

(3) $\frac{E_1}{E_2}\times 0.118$                                      

(4) $\frac{E_2-E_1}{0.118}$