Calculate the maximum work that can be obtained from the Daniell cell given below -

$Zn\left(s\right)<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>Z{n}^{2+}<mspace\; width="1em"></mspace>(aq)<mspace\; width="1em"></mspace>||<mspace\; width="1em"></mspace>C{u}^{2+<mspace\; width="1em"></mspace>}(aq)<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>Cu\left(s\right).$

Given that $E_{Z{n}^{2+}/Zn}^{°}=-0.76<mspace\; width="1em"></mspace>V<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>E_{C{u}^{2+}/Cu}^{°}=+0.34<mspace\; width="1em"></mspace>V$

(1) – 212300 J                                        

(2) – 202100 J

(3) – 513100 J                                        

(4) – 232120 J

 

The metal that cannot be produced on reduction of its oxide by aluminium is :

(1) K                                                        

(2) Mn

(3) Cr                                                      

(4) Fe

We have taken a saturated solution of AgBr, K_sp of AgBr is 12 × 10^-14.   If 10^-7 mole of AgNO₃  is added to 1 litre of this solution then the conductivity of this solution in terms of  10^-10   Sm^-1 units will be:

$[\lambda {°}_{\left(A{g}^{+}\right)}=6\times {10}^{-3}S{m}^{2}mo{l}^{-1}<mspace\; width="1em"></mspace>;$
$\lambda {°}_{\left(B{r}^{-}\right)}=8\times {10}^{-3}S{m}^{2}mo{l}^{-1},$
$\lambda {°}_{\left(N{O}_3^{-}\right)}=7\times {10}^{-3}S{m}^{2}mo{l}^{-1}]$

1. 39 
2. 55
3. 15 
4. 41

\(Z n →Z n^{2+}+2 e ; E^{\circ}=+0.76 \mathrm{~V}\)
\(F e → F e^{2+}+2 e ; E^{\circ}=+0.41 \mathrm{~V}\)

1. – 0.35 V                                            

2. + 0.35 V

3. + 1.17 V                                           

4. – 1.17 V

Which of following cell can produce more electrical work.

1. $$pt,H2|NH4Cl||0.1MCH3COOH|H2,pt

2. $pt,<mspace\; width="1em"></mspace>H_2<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>0.1<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>HCl<mspace\; width="1em"></mspace>||<mspace\; width="1em"></mspace>0.1<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>NaOH<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>H_2<mspace\; width="1em"></mspace>,<mspace\; width="1em"></mspace>pt$

3. $pt,<mspace\; width="1em"></mspace>H_2<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>0.1<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>HCl<mspace\; width="1em"></mspace>||<mspace\; width="1em"></mspace>0.1<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>C{H}_{3}COOK<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>H_2<mspace\; width="1em"></mspace>,<mspace\; width="1em"></mspace>pt$

4. $pt,<mspace\; width="1em"></mspace>H_2<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>0.1<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>C{H}_{3}COOK<mspace\; width="1em"></mspace>||<mspace\; width="1em"></mspace>0.1<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>HCl<mspace\; width="1em"></mspace>|<mspace\; width="1em"></mspace>H_2<mspace\; width="1em"></mspace>,<mspace\; width="1em"></mspace>pt$

The calomel electrode is reversible with respect to-

1. Mercury                                       

2. $H^{+}$

3. $H{g}^{2+}$                                          

4. $C{l}^{-}$

Zn amalgam is prepared by electrolysis of aqueous $ZnC{l}_2$ using Hg cathode (9 gm). How much current is to be passed through $ZnC{l}_2$ solution for 1000 seconds to prepare a Zn Amalgam with 25% Zn by wt. (Zn = 65.4)

(1) 5.6 amp 

(2) 7.2 amp

(3) 8.85 amp 

(4) 11.2 amp

The standard oxidation potentials of $Cu/C{u}^{2+}$and $C{u}^{+}<mspace\; width="1em"></mspace>/C{u}^{2+}<mspace\; width="1em"></mspace>are<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.34V<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.16<mspace\; width="1em"></mspace>V$respectively. The standard electrode potential of $C{u}^{+}<mspace\; width="1em"></mspace>/Cu$ would be :

1. 0.18V  

2. 0.52V

3. 0.82V    

4. 0.49V

Acidified water is electrolyzed using an inert electrode. The volume of gases liberated at

STP is 0.168L. The quantity of charge passed through the acidified water would be:

(1) 96,500 C 

(2) 9,650 C

(3) 965 C 

(4) 168 C

 ${\mathrm{E}}^0<mspace\; width="1em"></mspace>\mathrm{for}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{reaction}<mspace\; width="1em"></mspace>$
$\mathrm{Fe}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Zn}}^{2+}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\mathrm{Zn}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Fe}}^{2+}<mspace\; width="1em"></mspace>\mathrm{is}<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.35<mspace\; width="1em"></mspace>\mathrm{V}.$
The   given   cell   reaction   is: 

1. Feasible  

2. Not feasible

3. In equilibrium 

4. None of the above