Chemistry M.C.Q (1 Marks)

Mass of Cu deposited

$(w) $$ =\frac{M \times i \times t}{n F} $

$ =\frac{63 \times 9.6487 \times 100}{2 \times 96487} $

$ =0.315 \mathrm{~g}$

for 2 mole of $\mathrm{H}_2 \mathrm{O}=4 \mathrm{~F}$ charge is required for 1 mole of $\mathrm{H}_2 \mathrm{O}=\frac{4 \mathrm{~F}}{2}=2 \mathrm{~F}$ required

$\stackrel{+7}{\mathrm{Mn}} \mathrm{O}_4^{-} \rightarrow \stackrel{+2}{\mathrm{Mn}^{2+}}$

for $1$ mole $\mathrm{MnO}_4^{-} 5 \mathrm{~F}$ charge is required

$\mathrm{Ca}^{2+} \xrightarrow{+2 \mathrm{e}^{-}} \mathrm{Ca}$

For $1$ mole $\mathrm{Ca}^{2+}$ ion required $=2 \mathrm{~F}$

$1.5$ mole $\mathrm{Ca}^{2+}$ ion required $=\frac{2}{1} \times 1.5=3 \mathrm{~F}$

$\stackrel{+2}{\mathrm{FeO}} \rightarrow \stackrel{+3}{\mathrm{Fe}_2} \mathrm{O}_3$

for $1$ mole $\mathrm{FeO}, $1$ \mathrm{~F}$ charge is required.

$k =G G^*$

$=\frac{1}{R} G^*$

$G^* =k \times R=0.0210 \times 60=1.26\,cm ^{-1}$

$E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{0.059}{ n } \log \frac{\left[ Ni ^{+2}\right]^{1}}{\left[ Ag ^{+}\right]^{2}}$

$E _{\text {cell }}=10.5-\frac{0.059}{2} \log \frac{10^{-3}}{\left(10^{-3}\right)^{2}}$

$=10.5-\frac{0.059}{2} \log 10^{.3}$

$=10.5-\frac{0.059}{2} \times 3$

$=10.4115 \,V$

$\underset{ E _{ RP }^{\circ}=1.510 V }{2 MnO _{4}^{-}}+16 H ^{+}+10 e^{-} \rightarrow 2 Mn ^{+2}+8 H _{2} O$

Anode :

$5 H _{2} O \rightarrow \frac{5}{2} O _{2}+10 H ^{+}+10 e ^{-}$

$E _{ OP }^{\circ} =-1.223 \,V$

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Target reaction :

$2 MnO _{4}^{-}+ 6 H ^{+} \rightarrow 2 Mn ^{+2}+\frac{5}{2} O _{2}+3 H _{2} O$

$E _{\text {cell }}^{\circ} =( SRP )_{\text {Cathode }}-( SRP )_{\text {Andede }}$

$E _{ Cell }^{o} =1.510 \,V -1.223 \,V$

$=0.287 \,V$

Yes the given cell reaction is possible.

Reactivity order: $Zn \,>\, Fe\, >\, Cu\, >\, Ag$

In case of displacement reaction, more reactive metals (lower SRP) can displace less reactive metals (higher SRP) from their salt solution.

$CuSO _{4( aq )}+2 Ag _{( s )} \rightarrow Cu _{( s )}+ Ag _{2} SO _{4( aq )}$

Reaction is not possible as $Ag$ is less reactive metal compare to $Cu$.

$=400$

$\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\infty}}=\frac{20}{400}=0.05$

$=\mathrm{K}_{\mathrm{a}}=\mathrm{C} \alpha^{2}$

$=0.007 \times(0.05)^{2}$

$=7 \times 10^{-3} \times 25 \times 10^{-4}$

$=175 \times 10^{-7}$

$=1.75 \times 10^{-5}$

$\mathrm{HCl} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}....(2)$

$\mathrm{CH}_{3} \mathrm{COONa} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{Na}^{+}....(3)$

$\quad(2)+(3)-(1)$

$426.16+91-126.45$

$=390.71$

As per electrochemical series, $Au ^{3+}$ occupies. the top position.

At Anode : $2 H _{2} O \rightarrow O _{2( g )}+4 H _{( aq )}^{+}+4 e ^{-}$

Oxygen gas will liberate at anode

$2 H _{2}( g )+ O _{2}( g ) \longrightarrow 2 H _{2} O (\ell)$

$v.f.$ $=2$

As per faraday's $1^{\text {st }}$ law

Charge passed in faraday $=$ $g.$eq of product

$=\frac{20}{40} \times 2=1 F$

$K>A l>C r>A g$

$\mathrm{n}=2$

$\Delta G^{\circ}=-n F E^{\circ}$

$=-2 \times 96500 \times(0.24)$

$=-46320 \mathrm{J}$

$=-46.32 \mathrm{kJ} \mathrm{mol}^{-1}$

at equllbrium $\mathrm{E}_{\mathrm{cell}}=0, \mathrm{Q}=\mathrm{K}_{\mathrm{eq}}$

$0=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0591}{1} \log _{10} \mathrm{K}_{\mathrm{eq}}$

$\mathrm{E}_{\text {cell }}^{\circ}=+0.0591 \log _{10} \mathrm{K}_{\mathrm{eq}}$

$0.59=+0.0591 \log _{10} \mathrm{K}_{\mathrm{eq} .}$

$+10=\log _{10} \mathrm{K}_{\mathrm{eq}}$

$\mathrm{K}_{\mathrm{eq}}=10^{+10}$

$\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_{4}^{-2}\dots (2)$

$\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{K}^{+}+\mathrm{SO}_{4}^{-2}\dots (3)$

$\mathrm{CH}_{3} \mathrm{COOK} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{K}^{+}\dots (4)$

According to Kohlrausch's law

$\lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\circ}=\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{\circ}+\lambda_{\mathrm{H}^{+}}^{\circ}$

eq. $(1)=$ eq. $(4)+$ eq. $\frac{(2)}{2}-$ eq. $\frac{(3)}{2}$

$\therefore \quad \lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\circ}=z+\frac{\mathrm{x}}{2}-\frac{\mathrm{y}}{2}$

$\lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\circ}=\frac{(\mathrm{x}-\mathrm{y})}{2}+\mathrm{z}\left(\mathrm{S} \times \mathrm{cm}^{2} \mathrm{mol}^{-1}\right)$

$\mathrm{HBrO} \longrightarrow \mathrm{Br}_{2} \quad \mathrm{E}^{\circ}=1.595,$ SRP (cathode)

$\mathrm{HBrO} \longrightarrow \mathrm{BrO}_{3}^{-} \quad \mathrm{E}^{\circ}=-1.5 \mathrm{V},$ SOP (Anode)

$2 \mathrm{HBrO} \longrightarrow \mathrm{Br}_{2}+\mathrm{BrO}_{3}^{-}$

$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{SRP}(\text { cathode })-\mathrm{SRP}(\text { Anode })$

$=1.595-1.5$

$=0.095 \mathrm{V}$

$\mathrm{E}_{\text {cell }}^{\circ}>0 \Rightarrow \Delta \mathrm{G}^{\circ}<0$ [spontaneous]

$\Delta G ^{0}=- nF E _{ cell }^{0}$

$=-2 \times 96500 \times 1.10=-213.072\, k\,J / mol$

$=133.4+149.9-144.9$

$=138\, S \,cm ^{2}\, mol ^{-1}$

$\mathrm{Zn}\left|\mathrm{ZnSO}_{4}(0.01 \mathrm{M})\right||  \mathrm{CuSO}_{4}(1 \mathrm{M}) | \mathrm{Cu}$

Cell reaction $\rightarrow \mathrm{Zn}+\mathrm{Cu}^{+2} \longrightarrow \mathrm{Zn}^{+2}+\mathrm{Cu}$

${\mathrm{E}_{1}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \frac{\mathrm{Zn}^{+2}}{\mathrm{Cu}^{+2}}} $

${\mathrm{E}_{1}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \frac{0.01}{1}} $

${=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \frac{1}{100}}$

For cell

$\mathrm{Zn}\left|\mathrm{ZnSO}_{4}(1 \mathrm{M})\right|\left|\mathrm{CuSO}_{4}(0.01 \mathrm{M})\right| \mathrm{Cu}$

$\mathrm{E}_{2}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \frac{1}{0.01}$

$=\mathrm{E}^{\circ}-\frac{0.059}{2} \log 100 \ldots(2)$

$E_1 > E_2$

$\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}}^{0}=-0.76 \mathrm{V}$

$Zn$ has higher negative $SRP$ (Standard reduction potential) so it works as anode and protect iron to make iron as cathode.

$E_{H^{+} / H_{2}}=-\frac{0.0591}{2} \log \frac{P_{M_{2}}}{\left[H^{+}\right]^{2}}$

$\log \frac{P_{H_{2}}}{\left[H^{+}\right]^{2}}=0,$$ \frac{P_{H_{2}}}{\left[H^{+}\right]^{2}}=10^{0}=1$

$P H_{2}=\left[H^{+}\right]^{2}$

For pure $H_{2} O ; H^{+}=10^{-7} M$

$P_{H_{2}}=\left(10^{-7}\right)^{2}=10^{-14} \mathrm{atm}$

$=\frac{5.76 \times 10^{-3}\; \mathrm{Scm}^{-1} \times 1000}{0.5 \mathrm{mol} \mathrm{cm}^{-3}}$

$=11.52 \;\mathrm{S} \mathrm{cm}^{2} \mathrm{mol}^{-1}$

$Q=1 \times 60=60 \mathrm{C}$

$\text { Now, } 1.60 \times 10^{-19} \mathrm{C} \equiv 1 \text { electron }$

$\therefore 60 \mathrm{C}=\frac{60}{1.6 \times 10^{-19}}=37.5 \times 10^{19}$

$=3.75 \times 10^{20} \text { electrons }$

$2 \mathrm{H}_{2} \mathrm{O} \stackrel{2 \mathrm{e}^{-}}{\longrightarrow} \mathrm{H}_{2}+2 \mathrm{OH}^{-}$

At anode:

$2 \mathrm{Cl}^{-} \stackrel{2 e^{-}}{\longrightarrow} \mathrm{Cl}_{2}+2 \mathrm{e}^{-}$

$\frac{W}{E}=\frac{I t}{96500}$

$0.1 \times 2=\frac{3 \times t(\text {sec})}{96500}$

$t=6433 \mathrm{sec}$

$t=107.2 \mathrm{min}$

$\sim 110 \mathrm{min}$

$\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{4}^{-}+e^{-}$

$0.1 \mathrm{mol}\quad \quad \quad \quad \;\;\quad1 \mathrm{mol}$

$Q=0.1 \times F=0.1 \times 96500 \mathrm{C}=9650 \mathrm{C}$

$O_{2}$ at $STP$

Thus, $5600 \mathrm{mL} O_{2}$ means $=\frac{5600}{22400} \mathrm{mol} O_{2}$

$=\frac{1}{4} m o l O_{2}$

$\therefore$ Weight of $\mathrm{O}_{2}=\frac{1}{4} \times 32=8 \mathrm{g}$

According to problem,

Equivalents of $\mathrm{Ag}=$ Equivalents of $\mathrm{O}_{2}$

$\begin{array}{l} {=\frac{\text {Weight of } A g}{\text {Equivalent wor}} \mathrm{Ag}} \\ {=\frac{\mathrm{W}_{O_{2}}}{\text { Equivalent weight of } O_{2}}} \end{array}$

$\frac{W_{A g}}{\frac{M_{g}}{1}}=\frac{W_{O_{2}}}{\frac{M_{0_{2}}}{4}}$

$\therefore \quad \frac{W_{A g}}{108} \times 1=\frac{8}{32} \times 4$

$\left[\because 2 H_{2} O \rightarrow O_{2}+4 H^{+}+4 e^{-}\right]$

$\Rightarrow \quad W_{A g}=108 g$

$=0.76+0.34=1.10 \mathrm{V} $

$1 atm$ $\quad$ (at $pH\; 10$)

If $\mathrm{pH}=10$

$\mathrm{H}^{+}=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10}$

From nernst equation, $\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{0}-\frac{0.0591}{2} \log \frac{\left[\mathrm{H}^{+}\right]^{2}}{\mathrm{p}_{\mathrm{H}_{2}}}$

For hydrogen electrode, $E_{\text {cell }}^{0}=0$

$E_{\text {cell }}=-\frac{0.0591}{2} \log \frac{\left(10^{-10}\right)^{2}}{1}$

$0.0591 \times \log 10^{10}$

$0.0591 \times 10=0.591\; V$

$2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\right), E^{\circ}=+1.51 \mathrm{V}$.... $(ii)$

Subtracting Eq. $(ii)$ from Eq. $(i)$, we get

${3 \mathrm{Mn}^{2+} \longrightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}}$

${E^{\circ}=-1.18-(+1.51)=-2.69 \mathrm{V}}$

since, the value of $E^{\circ}$ is -ve, therefore the reaction is non-spontaneous

$Cr_2O_7^{2-} +2OH^- \to 2CrO_4^{2-} +H_2O$

$Fe \to F{e^{2 + }} + 2e$ (anode reaction)

${O_2} + 2{H_2}O + 4e \to 4O{H^ - }$ (cathode reaction)

The overall reaction is $2Fe + {O_2} + 2{H_2}O \to 2Fe{(OH)_2}$

$Fe{(OH)_2}$ may be dehydrated to iron oxide $FeO$, or further oxidised to $Fe{(OH)_3}$ and then dehydrated to iron rust, $F{e_2}{O_3}$.
 

$A{l^{3 + }}/Al\,\,\,\,\,\,\,\,{E^o} = - 1.662$

$S{n^{2 + }}/Sn\,\,\,\,\,{E^o} = - 0.136$

$P{b^{2 + }}/Pb\,\,\,\,\,{E^o} = - 0.126$

In galvanizing action $Zn$ is coated over iron.

$-\,E_{T{i^{2 + }}\,|T{i^{3 + }}}^o\, = \,\,0.37\,V$                $E_{T{i^{3 + }}\,|T{i^{2 + }}}^o\, = \, - \,0.37\,V$

$E_{T{i^{2 + }}\,|T{i^{3 + }}}^o\, = \,\,0.37\,V$

$2 H _2 O +2 e ^{-} \rightarrow H _2+2 OH ^{-}$

$\Delta G ^{\circ}=- nF E _{\text {cell }}^{\circ} \ldots\ldots 1$

$\rightarrow$ Emf of cell and equilibrium constant $K$ is given by ,

$E ^{\circ}=\frac{0.0591}{ n } \log K \ldots\ldots 2$

$\rightarrow$ The value of $K$ gives the extent of the cell reaction. If the value of $K$ is large, the reaction proceeds to larger extent.

$\rightarrow$ For spontaneous reaction, $\Delta G ^{\circ}$ is negative and $K\, >\,1$.

$\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}\left(\mathrm{C}_{1}\right)+\mathrm{e}^{-}$

At cathode

$\mathrm{Ag}^{+}(0.1 \mathrm{M})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})$

$E=E^{\circ}-\frac{0.0591}{1} \log _{10} \frac{C_{1}}{0.1\,M}$

$\mathop {{\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4}({\text{s}})}\limits_S  \rightleftharpoons \mathop {2{\text{A}}{{\text{g}}^ + }}\limits_{2S}  + \mathop {{\text{CrO}}_4^{2 - }}\limits_S $

$\left[\mathrm{Ag}^{+}\right]=2 \mathrm{s}=10^{-11}$

$\mathrm{K}_{\mathrm{sp}}=4 \mathrm{s}^{3}=4\left(\frac{10^{-11}}{2}\right)^{3}$

$\mathrm{E}_{\mathrm{OP}} =\mathrm{E}_{\mathrm{OP}}^{\circ}-\frac{0.059}{2} \log \left[\mathrm{Fe}^{2+}\right] $

$=0.44-\frac{0.059}{2} \log [0.1]=0.4695\,\mathrm{V}$