The standard reduction potentials for $M{n}^{3+}/M{n}^{2+}$ and $Mn{O}_2<mspace\; width="1em"></mspace>/M{n}^{3+}$ couples in acid medium at $25°C$ are 1.50 V and 1.00 V respectively. The following reaction:

$2M{n}^{3+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2H_{2}O<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>M{n}^{2+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>Mn{O}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>4H^{+}<mspace\; width="1em"></mspace>is$

(1) reversible                                             

(2) non-spontaneous

(3) spontaneous                                         

(4) irreversible

$M{n}^{3+}/M{n}^{2+}$और $Mn{O}_2<mspace\; width="1em"></mspace>/M{n}^{3+}$ युग्म के लिए मानक अपचयन विभव अम्लीय माध्यम में $25°C$ पर क्रमशः 1.50 V और 1.00 V हैं। निम्नलिखित अभिक्रिया $2M{n}^{3+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2H_{2}O<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>M{n}^{2+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>Mn{O}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>4H^{+}$ है:

(1) उत्क्रमणीय 

(2) अस्वत:

(3) स्वत:

(4) अनुत्क्रमणीय

The $k=<mspace\; width="1em"></mspace>4.95<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-5}S<mspace\; width="1em"></mspace>c{m}^{-1}$ for a 0.00099 M solution. Calculate the reciprocal of the degree of dissociation of acetic acid, if ${\wedge }_m^0$ for acetic acid is 400 S $c{m}^{2}mo{l}^{-1}$ 

1. 7                                        

2. 8

3. 9                                        

4. 10

एक 0.00099 M विलयन के लिए $k=<mspace\; width="1em"></mspace>4.95<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-5}S<mspace\; width="1em"></mspace>c{m}^{-1}$ है। एसीटिक अम्ल के वियोजन की मात्रा के व्युत्क्रम परिणाम की गणना कीजिए, यदि एसीटिक अम्ल के लिए ${\wedge }_m^0$ 400 S $c{m}^{2}mo{l}^{-1}$ है। 

(1) 7

(2) 8

(3) 9    

(4) 10

We have taken a saturated solution of $AgBr,<mspace\; width="1em"></mspace>K_{sp}$ of AgBr is $12<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-14}.<mspace\; width="1em"></mspace>$If ${10}^{-7}$mole of $AgN{O}_3$ are added to 1 litre of this solution then the conductivity of this solution in terms of ${10}^{-7}$ $S{m}^{-1}$ units will be

$[\lambda {°}_{\left(A{g}^{+}\right)}=4\times {10}^{-3}S{m}^{2}mo{l}^{-1}<mspace\; width="1em"></mspace>;$
$\lambda {°}_{\left(B{r}^{-}\right)}=6\times {10}^{-3}S{m}^{2}mo{l}^{-1},$
$\lambda {°}_{\left(N{O}_3^{-}\right)}=5\times {10}^{-3}S{m}^{2}mo{l}^{-1}]$

(1) 39 

(2) 55

(3) 15 

(4) 41

हमने AgBr का एक संतृप्त लिया, AgBr की$<mspace\; width="1em"></mspace>K_{sp}$ $12<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-14}$ है। यदि $AgN{O}_3$ के ${10}^{-7}$mol इस विलयन के 1 लीटर में मिलाए जाते हैं तो इस विलयन की चालकता ${10}^{-7}$ $S{m}^{-1}$ इकाई के संदर्भ में होंगी

$[\lambda {°}_{\left(A{g}^{+}\right)}=4\times {10}^{-3}S{m}^{2}mo{l}^{-1}<mspace\; width="1em"></mspace>;$
$\lambda {°}_{\left(B{r}^{-}\right)}=6\times {10}^{-3}S{m}^{2}mo{l}^{-1},$
$\lambda {°}_{\left(N{O}_3^{-}\right)}=5\times {10}^{-3}S{m}^{2}mo{l}^{-1}]$

(1) 39

(2) 55

(3) १५

(4) 41

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

$Cu/C{u}^{2+}$और $C{u}^{+}<mspace\; width="1em"></mspace>/C{u}^{2+}<mspace\; width="1em"></mspace>\mathrm{के}<mspace\; width="1em"></mspace>\mathrm{मानक}<mspace\; width="1em"></mspace>\mathrm{ऑक्सीकरण}<mspace\; width="1em"></mspace>\mathrm{विभव}<mspace\; width="1em"></mspace>\mathrm{क्रमशः}<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.34V<mspace\; width="1em"></mspace>\text{और}<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.16<mspace\; width="1em"></mspace>V$ हैं। $C{u}^{+}<mspace\; width="1em"></mspace>/Cu$का मानक इलेक्ट्रोड विभव होगा:

(1) 0.18V  

(2) 0.52V

(3) 0.82V

(4)0.49V

In a concentration cell, $Zn/Z{n}^{2+}<mspace\; width="1em"></mspace>(1.0<mspace\; width="1em"></mspace>M)<mspace\; width="1em"></mspace>\left|\right|<mspace\; width="1em"></mspace>Z{n}^{2+}<mspace\; width="1em"></mspace>(0.15<mspace\; width="1em"></mspace>M)/Zn$ as the cell discharges,

(1) reaction proceeds to the right

(2) the two solutions approach each other in concentration

(3) no reaction takes place

(4) water gets decomposed

एक सांद्रता सेल $Zn/Z{n}^{2+}<mspace\; width="1em"></mspace>(1.0<mspace\; width="1em"></mspace>M)<mspace\; width="1em"></mspace>\left|\right|<mspace\; width="1em"></mspace>Z{n}^{2+}<mspace\; width="1em"></mspace>(0.15<mspace\; width="1em"></mspace>M)/Zn$ में, सेल विसर्जन के रूप में,

(1) अभिक्रिया दाईं ओर बढ़ती है

(2) दो विलयन सांद्रता में एक दूसरे से संपर्क करते हैं

(3) कोई अभिक्रिया नहीं होती है

(4) जल विघटित हो जाता है

The standard potentials, $E^{\circ }$, for the half reactions are as $Zn<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>Z{n}^{2+}+<mspace\; width="1em"></mspace>2e;<mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.76<mspace\; width="1em"></mspace>V$and $Fe<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>F{e}^{2+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2e<mspace\; width="1em"></mspace>;<mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.41$ V; the e.m.f. for the cell reaction, $F{e}^{2+}+<mspace\; width="1em"></mspace>Zn<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>Z{n}^{2+}+<mspace\; width="1em"></mspace>Fe<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>is-$

(1) – 0.35 V                                            

(2) + 0.35 V

(3) + 1.17 V                                           

(4) – 1.17 V

मानक विभव, $E^{\circ }$, अर्ध अभिक्रियाओं $Zn<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>Z{n}^{2+}+<mspace\; width="1em"></mspace>2e;<mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.76<mspace\; width="1em"></mspace>V$ और $Fe<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>F{e}^{2+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2e,<mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.41$V के रूप में है। सेल अभिक्रिया $F{e}^{2+}+<mspace\; width="1em"></mspace>Zn<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>Z{n}^{2+}+<mspace\; width="1em"></mspace>Fe$ के लिए EMF है:

(1) – 0.35 V                                            

(2) + 0.35 V

(3) + 1.17 V

(4) – 1.17 V

The standard electrode potentials (reduction) of $Pt/F{e}^{3+},<mspace\; width="1em"></mspace>F{e}^{2+}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>Pt/S{n}^{4+},<mspace\; width="1em"></mspace>S{n}^{2+}<mspace\; width="1em"></mspace>are<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.77<mspace\; width="1em"></mspace>V<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.15<mspace\; width="1em"></mspace>V$ respectively at ${25}^{0}C.$The standard EMF of the reaction $S{n}^{4+}+<mspace\; width="1em"></mspace>2F{e}^{2+}<mspace\; width="1em"></mspace>\to S{n}^{2+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2F{e}^{3+}<mspace\; width="1em"></mspace>is-$

(A) – 0.62 V                                        

(B) – 0.92 V

(C) + 0.31 V                                        

(D) + 0.85 V

${25}^{0}C<mspace\; width="1em"></mspace>\text{पर}$$Pt/F{e}^{2+},<mspace\; width="1em"></mspace>F{e}^{3+}<mspace\; width="1em"></mspace>\text{और}<mspace\; width="1em"></mspace>Pt/S{n}^{4+},<mspace\; width="1em"></mspace>S{n}^{2+}<mspace\; width="1em"></mspace>\text{के\hspace{0.17em}मानक\hspace{0.17em}इलेक्ट्रोड\hspace{0.17em}विभव\hspace{0.17em}(अपचयन)\hspace{0.17em}क्रमशः}+<mspace\; width="1em"></mspace>0.77<mspace\; width="1em"></mspace>V<mspace\; width="1em"></mspace>\text{और}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.15<mspace\; width="1em"></mspace>V<mspace\; width="1em"></mspace>\text{हैं,}<mspace\; width="1em"></mspace>$अभिक्रिया $S{n}^{4+}+<mspace\; width="1em"></mspace>2F{e}^{2+}<mspace\; width="1em"></mspace>\to S{n}^{2+}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2F{e}^{3+}$ का मानक विद्युत वाहक बल है-

(1) – 0.62 V                                        

(2) – 0.92 V

(3) + 0.31 V

(4) + 0.85 V

Acetic acid has K_a = 1.8 × 10^–5 while formic acid had K_a = 2.1 × 10^–4. What would be the magnitude of the emf of the cell

Pt(H₂) $\left|\begin{array}{l}0.1\mathrm{M}<mspace\; width="1em"></mspace>\mathrm{acetic}<mspace\; width="1em"></mspace>\mathrm{acid}+\\ 0.1\mathrm{M}<mspace\; width="1em"></mspace>\mathrm{sodium}<mspace\; width="1em"></mspace>\mathrm{acetate}\end{array}\right|\left|\begin{array}{l}0.1\mathrm{M}<mspace\; width="1em"></mspace>\mathrm{formic}<mspace\; width="1em"></mspace>\mathrm{acid}+\\ 0.1\mathrm{M}<mspace\; width="1em"></mspace>\mathrm{sodium}<mspace\; width="1em"></mspace>\mathrm{formate}\end{array}\right|$ Pt(H₂) at 25°C

(1) 0.0315 volt

(2) 0.0629 volt

(3) 0.0455 volt

(4) 0.0545 volt

एसीटिक अम्ल में K_a = 1.8 × 10^–5 होता है जबकि फार्मिक अम्ल में K_a = 2.1 × 10^–4 होता है। सेल के विद्युत वाहक बल का परिमाण क्या होगा

Pt(H₂) $\left|\begin{array}{l}0.1\mathrm{M}<mspace\; width="1em"></mspace>\text{एसीटिक\hspace{0.17em}अम्ल}+\\ 0.1\mathrm{M}<mspace\; width="1em"></mspace>\text{सोडियम\hspace{0.17em}एसीटेट\hspace{0.17em}}\end{array}\right|\left|\begin{array}{l}0.1\mathrm{M}<mspace\; width="1em"></mspace>\mathrm{फार्मिक}<mspace\; width="1em"></mspace>\mathrm{अम्ल}+\\ 0.1\mathrm{M}<mspace\; width="1em"></mspace>\text{सोडियम\hspace{0.17em}फार्मेट}<mspace\; width="1em"></mspace>\end{array}\right|$ 25°C पर Pt(H₂)

(1) 0.0315 volt

(2) 0.0629 volt

(3) 0.0455 volt

(4) 0.0545 volt

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

$\left(i\right)<mspace\; width="1em"></mspace>A+e^{-}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>A^{-}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.24$

$\left(ii\right)<mspace\; width="1em"></mspace>B^{-}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>e^{-}\to <mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>B^{-2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1.25$

$\left(iii\right)<mspace\; width="1em"></mspace>C^{-}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2^{e-}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>C^{-3}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E{<mspace\; width="1em"></mspace>}^{-}=<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>1.25$

$\left(iv\right)<mspace\; width="1em"></mspace>D<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2^{e-}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>D^{-2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>0.06$

$\left(v\right)<mspace\; width="1em"></mspace>E<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>4^{e-}\to E^{-4}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>0.38$

(A) i and ii                                                   

(B) ii and iii

(C) iv and v                                                 

(D) ii and v

कुछ अर्ध सेल अभिक्रिया और उनकी मानक विभव दी गयी है, जिसके संयोजन से सबसे ज्यादा विभव वाले सेल का परिणाम होगा।

$\left(i\right)<mspace\; width="1em"></mspace>A+e^{-}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>A^{-}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.24$

$\left(ii\right)<mspace\; width="1em"></mspace>B^{-}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>e^{-}\to <mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>B^{-2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1.25$

$\left(iii\right)<mspace\; width="1em"></mspace>C^{-}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2^{e-}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>C^{-3}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E{<mspace\; width="1em"></mspace>}^{-}=<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>1.25$

$\left(iv\right)<mspace\; width="1em"></mspace>D<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2^{e-}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>D^{-2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>0.06$

$\left(v\right)<mspace\; width="1em"></mspace>E<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>4^{e-}\to E^{-4}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>0.38$

(1) i और ii

(2) ii और iii

(3) iv और v

(4) ii और v

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

नीचे दिए गए डेनियल सेल से प्राप्त अधिकतम कार्य की गणना कीजिए-

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

दिया गया है कि $E_{Z{n}^{2+}/Zn}^{°}=-0.76<mspace\; width="1em"></mspace>V<mspace\; width="1em"></mspace>\text{और}<mspace\; width="1em"></mspace>E_{C{u}^{2+}/Cu}^{°}=+0.34<mspace\; width="1em"></mspace>V$

(1) – 212300 J                                        

(2) – 202100 J

(3) – 513100 J                                        

(4) – 232120 J