$F_2$ gas can’t be obtained by the electrolysis of any $F^{-1}$salt because-

(1) Fluorine is the strongest reducing agent

(2) Fluorine is the strongest oxidising agent.

(3) Fluorine easily combine with atmospheric $O_2$

(4) All

Calculate the quantity of electricity (i.e. charge) delivered by a Daniel cell initially containing 1L each of $1 M C{u}^{2+}$ion and $1M Z{n}^{2+},$which is operated until its potential drops to 1V.
(Given : \(E_{Zn2+/Zn}^0 = -0.76V; E_{Cu2+/Cu}^0 = +0.34V\))
(1) 0.029 x 10⁵C
(2) 2.239 x 10⁵C
(3) 1.92 x 10⁵C
(4) 3.123 x 10⁵C

Some half cell reaction & their standard potential are given which combination would result in a cell with the largest potential.

$\left(i\right) A+e^{-} \to A^{-} E° = – 0.24$

$\left(ii\right) B^{-} + e^{-}\to B^{-2} E° = 1.25$

$\left(iii\right) C^{-} + 2^{e-} \to C^{-3} E{ }^{-}= – 1.25$

$\left(iv\right) D + 2^{e-} \to D^{-2} E° = 0.06$

$\left(v\right) E + 4^{e-}\to E^{-4} E° = 0.38$

(A) i and ii                                                   

(B) ii and iii

(C) iv and v                                                 

(D) ii and v

A hydrogen electrode is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to the solution. $(Take p_{H_{2_{}}} = 1 atm) T = 298 K.$

(1) increase by 0.41 V                             

(2) increase by 59 mV

(3) decrease by 0.41 V                            

(4) decrease by 59 mV

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