The solubility product of silver iodide is $8.3 \times {10}^{-17}$and the standard potential (reduction) of Ag,$A{g}^{+}$ electrode is + 0.800 volts at ${25}^{°}C.$The standard potential of Ag, AgI/$I^{-}$ electrode (reduction) from these data is-

(1) – 0.30 V                            

(2) + 0.15 V

(3) + 0.10 V                            

(4) – 0.15 V

How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere-

1. 26.805 h                                   

2. 53.61 h

3. 107.22 h                                   

4. None of these

The standard redox potentials E^o of the following systems are-
 

The oxidizing power of the various species decreases in the order-

1. $C{e}^{4+}>C{r}_2{O_7}^{2-}> S{n}^{4+} > Mn{O_4}^{-}$

2. $C{e}^{4+}>Mn{O_4}^{- }>C{r}_2{O_7}^{2-}> S{n}^{4+}$

3. $C{r}_2{O_7}^{2-}>S{n}^{4+}>C{e}^{4+}+ Mn{O_4}^{-}$

4. $Mn{O_4}^{-}>C{e}^{4+} > S{n}^{4+}>C{r}_2{O_7}^{2-}$

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

$Zn\left(s\right) | Z{n}^{2+} (aq) || C{u}^{2+ }(aq) | Cu\left(s\right).$

Given that $E_{Z{n}^{2+}/Zn}^{°}=-0.76 V and E_{C{u}^{2+}/Cu}^{°}=+0.34 V$

(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:

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

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 || 0.1 M CH3COOH | H2 , pt

2. $pt, H_2 | 0.1 M HCl || 0.1 M NaOH | H_2 , pt$

3. $pt, H_2 | 0.1 M HCl || 0.1 M C{H}_{3}COOK | H_2 , pt$

4. $pt, H_2 | 0.1 M C{H}_{3}COOK || 0.1 M HCl | H_2 , pt$

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