Calculate the emf of the cell in which of the following reaction takes place:\$Ni(s) + 2Ag^+ (0.002 M) \rightarrow Ni^{2+} (0.160 M) + 2Ag(s)\$[Given that \$E^{\circ}_{cell} = 1.05 V\$]

Given,\$[Ag^+]=0.002M\$\$[Ni^{2+}]=0.160M\$\$n=2\$According to Nernst equation:\$E_{cell}=E^0-\dfrac{0.0591}{2}\log \dfrac{[Ni^{2+}]}{[Ag^+]^2}\$\$\therefore E_{cell}=1.05-\dfrac{0.0591}{2}\log \dfrac{0.160}{(0.002)^2}\$\$\therefore E_{cell}=1.05-0.02955\, \log{(4\times 10^4)}\$\$\therefore E_{cell}=0.91V\$Hence, the correct option is \$\text{B}\$

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Introduction to Electrolysis

Q: How would you determine the standard electrode potential of the \$Mg^{2+}\$ \$|Mg\$?

To determine the electrode potential of \$mg^{+2}/mg^{1}\$ we use hydrogen electrode (anode)The standard hydrogen electrode potential is zero.\$\therefore E^{o}cell = E^{o} right+ E^{o} left\$\$E^{o} cell= E^{o} mg^{+2}/mg^{-1} E^{o}n_{2}/n_{1}\$ \$E^{o} mg^{+2}/mg= E^{o} cell\$

Q: How much electricity in terms of Faraday is required to produce:(i) \$20\ g\$ of \$Ca\$ from molten \$CaCl_{2}\$(ii) \$40\ g\$ of \$Al\$ from molten \$Al_{2}O_{3}\$[Given : Molar mass of calcium & Aluminium are 40 g \$mol^{-1}\$ & 27 g \$mol^{-1}\$ respectively]

\$(1)\$ \$\displaystyle Ca^{2+} + 2e^- \rightarrow Ca\$\$1\$ mole (\$40\$ g) of \$Ca\$ will require \$2\$ F of electricity.\$20\$ g (\$0.5\$ mole) of \$Ca\$ will require \$1\$ F of electricity.(2) \$\displaystyle Al^{3+} + 3e^- \rightarrow Al \$\$1\$ mole (\$27\$ g) of \$Al\$ will require \$3\$ moles of electrons.\$40\$ g of \$Al\$ will require \$\displaystyle \dfrac {3 \times 40F}{27} = 4.4\$ F

Q: During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is:

Explanation:We have to use the formula \$W = \frac E {96500} \times I \times t\$.Consider formation of chlorine molecule as shown below:\$2Cl^- \rightarrow Cl_2 + 2e^-\$using formula,\$0.1 \times 71\$ (Molecular mass of \${Cl}_2\$) \$=\frac {35.5}{96500} \times 3 \times t\$.\$\therefore\$ \$t= 6433.33\ seconds \$\$\therefore\$ \$t\approx\$ \$110\ minutes\$Final Answer: Hence option \$D\$ is correct.

Q: A solution of \$Ni(NO_{3})_{2}\$ is electrolysed between platinum electrodes using a current of \$5\$ amperes for \$20\$ minutes. What mass of \$Ni\$ is deposited at the cathode?

Given\$I = 5A\$\$Time = 20 \times 60 = 1200\,\,s\$\$\therefore Charg e = current \times time\$\$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5 \times 1200\$\$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6000\,\,C\$According to the reaction.\$N{i^{2 + }}\left( {aq.} \right) + 2e \to Ni\,\left( s \right)\$\$Nickel\,\,deposite\,by\, (2 \times 96487)\,C = 58.7\,\,g\$\$\therefore Nickel\,deposite\,by\ 6000\,C = \dfrac{{58.7 \times 6000}}{{2 \times 96487}}\$\$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad= 1.825\,\,g\$\$\therefore \$ Hence \$1.825\,g\$ of nickel will be deposited at the cathode.

Q: Three electrolytic cells \$A, B, C\$ containing solutions of \$ZnSO_{4}, AgNO_{3}\$ and \$CuSO_{3}\$, respectively are connected in series. A steady current of \$1.5\$ amperes was passed through them until \$1.45\ g\$ of silver deposited at the cathode of cell \$B\$. How long did the current flow? What mass of copper and zinc were deposited?

Cell \$B\$: \$\displaystyle Ag^+ + e^- \rightleftharpoons Ag\$ at cathode.\$1\$ mole (\$108\$ g) of Ag is deposited by \$96500\$ \$C\$.\$1.45\$ g of Ag will be deposited by \$\displaystyle \dfrac {96500 \times 1.45}{108} = 1295.6 \: C\$.Now,\$\displaystyle Q = It \$\$\displaystyle 1295.6 = 1.5 \times t \$\$\displaystyle t = 864 \: s\$Cell \$A\$: \$\displaystyle Zn^{2+} + 2e^- \rightarrow Zn\$\$2\$ moles of electrons (\$\displaystyle 2 \times 96500\$ C of current) produces \$1\$ mole (\$63.5\$ g) of zinc.\$1295.6\$ \$C\$ of electricity will deposit \$\displaystyle \dfrac {65.3}{2 \times 96500} \times 1295.6 = 0.438\$ g of zincCell \$C\$: \$\displaystyle Cu^{2+} + 2e^- \rightarrow Cu\$\$2\$ moles of electrons (\$\displaystyle 2 \times 96500\$ \$C\$) of current will produce \$1\$ mole (\$63.5\$ g) of \$Cu\$. \$1295.6\$ \$C\$ of current will deposit \$\displaystyle \dfrac {63.5 \times 1295.6}{2 \times 96500} = 0.426 \: g\$ of copper

Q: How much charge is required for the following reductions?(i) \$1\ mol\$ of \$Al^{3+}\$ to \$Al\$(ii) \$1\ mol\$ of \$Cu^{2+}\$ to \$Cu\$(iii) \$1\ mol\$ of \${MnO_4}^{-}\$ to \$Mn^{2+}\$

\$Al^{3+} + 3e^-\rightarrow Al\$The reduction will require 3 Faraday of electricity or \$\displaystyle 3 \times 96500 = 2.895 \times 10^5 \: C\$.\$Cu^{2+} + 2e^-\rightarrow Cu\$The reduction will require 2 Faraday of electricity or \$\displaystyle 2 \times 96500 = 1.93 \times 10^5 \: C\$.\${MnO_4}^{-} + 5e^- \rightarrow Mn^{2+} \$The reduction will require 5 Faraday of electricity or \$\displaystyle 5 \times 96500 = 4.825 \times 10^5 \: C\$.Note: 1 mole of electron has a charge of 1 faraday.

Q: The weight of silver (atomic weight \$= 1 08\$) displaced by a quantity of electricity which displaces 5600 ml of \$O_2\$ at STP will be:

At STP, 1 mole of oxygen occupies 22400 mL.Hence, the number of moles of oxygen corresponding to 5600 ml is \$n_{O_2}=\dfrac {5600}{22400}=\dfrac {1}{4}=\dfrac {W_{O_2}}{M_{O_2}}\$.1 mole of oxygen produces 4 moles of lectrons and 1 mole of silver requires 1 mole of electrons.\$\dfrac {W_{Ag}}{M_{Ag}}\times 1=\dfrac {W_{O_2}}{M_{O_2}}\times 4\$ \$\dfrac {W_{Ag}}{M_{Ag}}\times 1=\dfrac {1}{4}\times 4\$ \$\dfrac {W_{Ag}}{108}=\dfrac {1}{4}\times 4\$\$W_{Ag}=108 g\$ Thus, the weight of silver displaced by a quantity of electricity which displaces 5600 ml of \$O_2\$ at STP will be 108.0 g.

Q: How much electricity is required in coulomb for the oxidation of:(i) \$1\ mol\$ of \$H_{2}O\$ to \$O_{2}\$?(ii) \$1\ mol\$ of \$FeO\$ to \$Fe_{2}O_{3}\$?

(i) \$\displaystyle H_2O \rightarrow 2H^+ + \dfrac{1}{2}O_2 + 2e^-\$Oxidation of one mole of water will require \$\displaystyle 2 \times 96500 = 1.93 \times 10^5 \: C\$.(ii) \$\displaystyle Fe^{2+} \rightarrow Fe^{3+} \$Oxidation of one mole of \$\displaystyle Fe^{2+} \$ will require \$\displaystyle 1 \times 96500 = 96500 \: C\$\$\displaystyle \$.

Q: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Q=ItWhere Q= no. of electronI= currentt= time taken\$Q=0.5 A\times 72005\$\$=3600 C\$As \$96487 C = 6.023\times 10^{23}\$ no. of electron\$\Rightarrow 3600 C = \frac{6.023\times 10^{23}\times 3600}{96487}=2.25\times 10^{22}\$ no. of electron

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