 # Electrochemistry

## Chemistry

### NCERT

1   How would you determine the standard electrode potential of the system $Mg^{2+1} Mg$?

A cell will be set up consisting of $Mg/MgSO_4 (1 M)$ as one electrode and standard hydrogen electrode $Pt, H, (1 atm)H ^+ /(l M)$ as second electrode, measure the emf of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that $e ^{-1}$ s flow from Mg electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as follows: $Mg|Mg^{2+} (1 M) || H ^+ (1 M) | H _2 , (1 atm) Pt$$\\ E _{cell} = E ^o _{H^+\frac{1}{2}H_2}-E^oMg^{2+}/Mg$$\\$ Put $E^o_{H+\dfrac{1}{2}H_2}=0$$\\ \therefore E^o_{Mg^{2+}/Mg}=-E^o_{cell} 2 Can you store copper sulphate solutions in a zinc pot? ##### Solution : Zn being more reactive than Cu, displaces Cu from CuSO_4 solution as follows:\\ Zn_{(s)} + CuSO_{4 (aq)} \to ZnSO4_{(aq)} +Cu_{(s)}$$\\$ In terms of emf, we have$\\$ $Zn|Zn ^{2+} || Cu^{ 2+} | Cu$$\\ E^o_{cell}=E^o_{Cu^{2+}/Cu} -E^o_{Zn^{2+}/Zn}$$\\$ $=0.34 V-(-0.76V)=1.10V$$\\ As E^o_{cell} is positive, reaction takes place, i.e., Zn reacts with copper and hence, we cannot store CuSO4 Solution in zinc pot.\\ 3 Consult the table of standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions. ##### Solution : Oxidation of Fe^{2+} converts it to Fe^{3+} , i.e., Fe^{ 2+} \to Fe ^{3+ }+e^{-} ; E^o_{ OX} = - 0.77 V Only those substances can oxidise Fe^{ 2+} to Fe^{ 3+} which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V, so that emf of the cell reaction is positive. This is so for elements lying below Fe ^{3+} /Fe ^{2+} in the series ex: Br_ 2 , Cl_ 2 and F_2 4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. ##### Solution : For hydrogen electrode, H^+ + e^ - \to 1/2 H_2$$\\$ Applying Nernst equation,$\\$ $E_{H^{-\frac{1}{2}}H_2} = E^o_{H^\frac{1}{2}H_2}-\dfrac{0.0591}{n}log\dfrac{1}{[H^+]}$$\\ =0-\dfrac{0.0591}{1}log\dfrac{1}{10^{-10}}$$\\$ $\{pH=10 \Rightarrow [H^+]=10^{-10}M \}$ $\\$ $=-0.0519*10$$\\ =0.591 V 5 Calculate the emf of the cell in which the following reaction takes place: Ni_{(s)} +2Ag^+ (0.002 M) \to Ni^{2+} (0.160 M)+2Ag_{(s)} Given that = E^o_{(cell)} 1.05 V. ##### Solution : Applying Nernst equation\\ E_{cell}=E^o_{cell}-\dfrac{0.0591}{n}\log \dfrac{[Ni^{2+}]}{[Ag^+]^2}\\ =1.05V-\dfrac{0.0519}{2}\log \dfrac{0.160}{(0.002)^2}\\ =1.05-\dfrac{0.0591}{2}\log (\log 4 * 10^{4})\\ =1.05 -\dfrac{0.0591}{2}(4.6021)\\ =1.05-0.14V=0.91 V$$\\$

6   Calculate the emf of the cell in which the following reaction takes place:$\\$ $Ni _{(s)} +2Ag^ + (0.002 M) \to Ni^{ 2+} (0.160 M)+2Ag_{ (s)}$ Given that = $E ^o_{ (cell)} 1.05 V.$

##### Solution :

Applying Nernst equation $\\$ $E_{cell}=E^o _{cell}-\dfrac{0.0591}{n}\log \dfrac{[Ni^{2+}]}{[Ag^+]^2}$ $\\$ $=1.05v-\dfrac{0.0519}{2}\log \dfrac{0.160}{(0.002)^2}$ $\\$ $=1.05-\dfrac{0.0591}{2}\log (\log4*10^4)$ $\\$ $=1.05-\dfrac{0.0591}{2}(4.6021)$ $\\$ $=1.05-0.14V=0.91V$

7   The cell in which the following reaction occurs: $2Fe ^{3+}_{(aq)} + 21^ -_{(aq)} \to 2Fe^{ 2+}_{(aq)} + l_{ 2(s)}$ has $E^o _{cell} = 0.236 V$ at $298 K$ . calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

##### Solution :

$2 Fe ^{3+}_{(aq)} + 2I^ -_{ (aq)} \to 2Fe ^{2+}_{(aq)} + I_{ 2 (s)}$ $\\$ For the given cell, $n = 2$ $\\$ $\Delta _ r G =- nF E ^o_{ cell}$ $\\$ $= -2 * 96500 * 0.236$ $\\$ $= -45.55 kJ$ mol $^{ -1}$ $\\$ Also, $\Delta _ r G^o = -2.303$ RT $\log KC$ $\\$ $\Rightarrow \log K_C=\dfrac{\Delta _rG^o}{2.303RT}=\dfrac{-45.55}{2.303*8.314*10^{-3}*298}=7.983$ $\\$ $\rightarrow K_C=$ antilog$(7.983)$ $\\$ $=9.616*10^7$

8   Why does the conductivity of solution decrease with dilution?

##### Solution :

Conductivity of a solution is the conductance of ions present in a unit volume of the solutions. On dilution, number of ions per unit volume decreases. Hence, the conductivity decreases.

9   Suggest a way to determine the value $\Lambda^o_ m$ of water.

##### Solution :

By using Kohlrausch’s law,$\Lambda^o_ m$ for $H_ 2 O$ can be calculated, we can write,$\\$ $\Lambda^o_ m= \Lambda^o_ m(Hcl)+\Lambda^o_ m(NaOH)-\Lambda^o_ m (NaCl)$ $\\$ Being strong electrolytes,$\Lambda^o_ m$ values of $HCl,NaOH$ and $NaCl$ are known $\\$ By substituting their values, we can obtain $\Lambda^o_ m$ for $H_2O$

10   The molar conductivity of $0.025$mol $L ^{-1}$ methanoic acid is $46.1 S cm ^2$ mol$^{ -1}$ . Calculate its degree of dissociation and dissociation constant Given $\lambda ^ 0 ( H ^+ ) = 349.6 cm ^2$ and $\lambda^o (HCOO ^- ) = 54.6 cm ^2 mol^{ -1}$

##### Solution :

$\Lambda ^o_ m ( HCOOH ) =\lambda ( H ^+ ) +\lambda ^o( HCOO ) ^-$ $\\$ $= 349.6 + 54.6$ $\\$ $= 404.2 S cm ^2$ mol$^{ -1}$ $\\$ $\Lambda ^C _m = 46.1 S cm ^2$ mol$^{ -1}$ $\\$ $\therefore \alpha \therefore \dfrac{\Lambda^C_m}{\Lambda^o_m}=\dfrac{46.1}{404.2}=0.114$ $\\$ $HCOOH\rightleftharpoons HCOO^- + H ^+$ $\\$ Initial conc.at equi,$c(1-\alpha )\qquad c\alpha \qquad c\alpha$ $\\$ $\therefore K_a=\dfrac{c\alpha.c\alpha}{c(1-\alpha)}=\dfrac{c\alpha^2}{1-\alpha}\\ =\dfrac{0.025*(0.114)^2}{1-0.114}=3.67*10^{-4}$

11   Suggest a list of metals that are extracted electrolytically.

##### Solution :

Na, Ca, Mg, and Al

12   Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

##### Solution :

From the reaction 1 mol of $Cr _2 O _7^{2-}$ require 6F$\\$ $= 6 *96500 = 579000 C$$\\ \therefore 579000 C of electricity are required for reduction of Cr _2 O_ 7^{2-} to Cr ^{3+} A lead storage battery consists of anode of lead, cathode of a gird of lead packed with lead dioxide (PbO _2 ) and 38% H _2 SO_ 4 solution as electrolyte. When the battery is in use, the reaction taking place are:\\ Anode : Pb ( s ) + SO_ 4^{ 2- } ( aq ) \to PbSO _4 ( s ) + 2 e ^-$$\\$ Cathode : $PbO _2 ( s ) + SO_ 4 ^{2 -} ( aq ) + 4 H ^+ ( aq ) \to 2 e ^-+ PbSO_ 4 ( s ) + 2 H _2 O ( l )$$\\ Over all reaction : Pb ( s ) + PbO_ 2 ( s ) + 2 H _2 SO_ 4 ( aq ) \to 2 PbSO _4 ( s ) + 2 H_ 2 O ( l )$$\\$ On charging the battery, the reverse reaction takes place, i.e., PbSO 4 deposited on electrodes is converted back to $Pb$ and $PbO _2$ and $H _2 SO_ 4$ is regenerated.

13   Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

##### Solution :

Methane and Methanol.

14   Explain how rusting of iron is envisaged as setting up of an electro chemical cell

##### Solution :

The Water present on the surface of iron dissolves acidic oxides of air like $CO _2 , SO _2$ etc, to form acids which dissociate to give $H ^+$ ions:$\\$ $H _2 O + CO _2 \to H_ 2 CO_ 3 \leftrightarrow 2H^ + + CO_ 3^{2-}$$\\ In the presence of H ^+ , iron loses e^{ -1} to form Fe ^{3+} .\\ Hence, this spot acts as anode:\\ Fe (s) \to Fe^{ 2+}(aq) + 2e^ -$$\\$ The $e^{ -1}$ released move through the metal to reach another spot, where $H ^+$ ions and dissolved oxygen take up these e -1 s and reduction occurs. This spot, thus acts as cathode:$\\$ $O_ 2 (g) + 4H ^+ (aq) +4e^ - \to 2H_ 2 O (l)$$\\ The overall reaction is:\\ 2Fe (s) + O _2 (g) + 4H^ +(aq) \to 2Fe^{ 2+}(aq) + 2H_ 2 O (l)$$\\$ Thus, an electro chemical cell is set up on the surface. Ferrous ions are further oxidized by atmospheric oxygen to ferric ions which combine with water to form hydrated ferric oxide, $Fe _2 O_ 3 . XH _2 O$which is rust.

15   Depict the galvanic cell in which the reaction,$\\$ $Zn(s) + 2Ag ^+ (aq) \to Zn ^{2+} (aq) + 2Ag(s)$ takes place. Further show:$\\$ (i) Which of the electrode is negatively charged?$\\$ (ii) The carriers of the current in the cell.$\\$ (iii) Individual reaction at each electrode.$\\$

##### Solution :

Mg, Al, Zn, Fe, Cu, Ag

The set-up will be similar to as shown below,

Higher the oxidation potential more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be Ag < Hg < Cl < Mg < K.

##### Solution :

(i) Cell reaction:$\\$ $Mg + Cu^{ 2+} \to Mg^{ 2+} + Cu (n = 2)$$\\ Nernst equation:\\ E_{cell}=E^o_{cell}-\dfrac{0.0591}{2}\log \dfrac{[Mg^{2+}]}{[Cu^{2+}]}\\ \therefore E_{cell}=0.34-(-2.37)-\dfrac{0.0591}{2}\log \dfrac{10^{-3}}{10^{-4}}\\ =2.71 - 0.02955 = 2.68 V (ii) Cell reaction:\\ Fe + 2H ^+ \to Fe^{ 2+ }+ H _2 (n = 2)\\ E_{cell}=E^o_{cell}-\dfrac{0.0591}{2}\log \dfrac{[Fe^{2+}]}{[H^+]^2}\\ \therefore E_{cell}=0-(-0.44)-\dfrac{0.0591}{2}\log \dfrac{10^{-3}}{(1)^2}\\ =0.44-\dfrac{0.0591}{2}*(-3)\\ =0.44+0.0887=0.5287 V \\ (iii) Cell reaction:\\ Sn + 2H^ + \to Sn^{ 2+} + H _2 (n = 2) Nernst equation:\\ E_{cell}=E^o_{cell}-\dfrac{0.0591}{2}\log \dfrac{sn^{2+}}{[H^+]^2}\\ E_{cell}=E^o_{cell}-\dfrac{0.0591}{2}\log \dfrac{0.05}{(0.02)^2}\\ =0-(-0.14)-\dfrac{0.0591}{2}\log \dfrac{0.05}{(0.02)^2}\\ =0.14-\dfrac{0.0591}{2}\log 125\\ =0.14-\dfrac{0.0591}{2}(2.0969)=0.078V \\ (iv) Cell reaction:\\ 2Br^ - + 2H ^+ \to Br _2 + H _2 (n = 2)$$\\$ Nernst equation:$\\$ $E_{cell}=E^o_{cell}-\dfrac{0.0591}{2}\log \dfrac{1}{[Br^-]^2[H^+]^2}\\ =(0-1.08)-\dfrac{0.0591}{2}\log \dfrac{1}{(0.01)^2(0.03)^2}\\ =-1.08-\dfrac{0.0591}{2}\log(1.111*10^7)\\ =-1.08-\dfrac{0.0591}{2}(7.0457)\\ =-1.08-0.208=-1.288 V.$ $\\$ Thus, oxidation will occur at the hydrogen electrode, and reduction will occur on $Br _2$ electrode.

18   Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

##### Solution :

Zn is oxidized and $Ag _2 O$ is reduced.$\\$ $E ^o_{ cell} = E ^o_{ Ag _2 O , Ag ( reduction )} - E ^o_{ Zn / Zn ^{2 +} ( oxidation )}\\ = 0.344 + 0.76 = 1.104 V\\ \Delta G^o = nFE^o _{cell} = -2 * 96500 * 1.104 J\\ = -2.13 * 10 ^5 J.$

Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.$\\$ Molar conductivity of a solution at a dilution (V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm 3 of the solution when the electrodes are one cm apart and the area of cross-section of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by $\land _m$.$\\$ The conductivity of a solution (both for strong and weak electrolytes) decreases with decrease in concentration of the electrolyte, i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases with decrease in concentration of the electrolyte, i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases with decrease in concentration of the electrolyte. This is because both number of ions as well as mobility of ions increases with dilution. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

19   The resistance of a conductivity cell containing $0.001 M KCl$ solution at $298 K$ is $1500 \omega$ .What is the cell constant if conductivity of $0.001 M KCl$ solution at $298 K$ is $0.146 * 10 ^{-3} S cm^{ -1} ?$

##### Solution :

$\land _m=\dfrac{k*1000}{\text{Molarity }}=\dfrac{0.0248\ S\ cm^{-1}*1000 cm^3\ L^{-1}}{0.20\ mol \ L^{-1}}\\ =124S \ cm^2\ mol^{-1}$

Cell constant =$\dfrac{\text{Conductivity}}{\text{Conductance}} \\ =\text{Conductivity}* \text{Resistance}\\ =0.146 * 10^{ -3} S cm ^{-1} * 1500 \omega \\ =0.218 \ cm^{-1}$

##### Solution :

$\land ^c_m=\dfrac{K*1000}{\text{Molarity}}\\ \dfrac{(7.896*10^{-1})*1000 cm^3 L^{-1}}{0.00241 mol\ L^{-1}}\\ =32076 S cm^2 mol^{-1}\\ \alpha =\dfrac{\land^o_m}{\land^o_m}-\dfrac{32.76}{390.5}=8.4* 10^{-2}\\ k_a=\dfrac{C\alpha^2}{1-\alpha}=\dfrac{0.0024*(804*10^{-2})^2}{1-0.084}=1.86*10^{-5}$

(i) $Ca^{ 2+} + 2e ^- \to Ca$$\\ Thus, 1 mol of Ca, i.e., 40g of Ca require = 2F electricity \therefore 20 g of Ca require = 1 F of electricity\\ (ii) Al ^{3+} + 3e^ - \to A1$$\\$ Thus, 1 mole of Al, i.e., 27g of Al require = 3 F electricity $\therefore$ 40g of Al will require electricity$\\$ $=\dfrac{3}{27}*40=4.44F$ of electricity.

(i) The electrode reaction is $Al ^{3+} + 3e \to Al$$\\ \therefore Quantity of charge required for reduction of 1 mol of Al ^{3+} = 3F = 3 * 96500 C = 289500 C.$$\\$ (ii) The electrode reaction is $Cu^{ 2+} + 2e ^- \to Cu$$\\ \therefore Quantity of charge required for reduction of 1 mol of Cu^{ 2+} = 2F = 2 * 96500 = 193000 C.$$\\$ (iii) The electrode reaction is $MnO _4^- \to Mn ^{2+}$ i.e., $Mn ^{7+} + 5e^ - \to Mn^{ 2+}$$\\ \therefore Quantity of charge required = 5F = 5 * 96500 C = 4825000 C. 22 How much electricity is required in coulomb for the oxidation of\\ (i) 1 mol of H _2 O to O _2$$\\$ (ii) 1 mol of $FeO$ to $Fe_ 2 O_ 3$$\\ ##### Solution : (i) The electrode reaction for 1 mol of H _2 O is \\ H_2O \to H_2+\dfrac{1}{2}O_2\\ i.e.,O^{2-}\to \dfrac{1}{2}O_2+2E^- \\ \therefore Quantity of electricity required\\ =2F = 2 * 96500 C = 193000 C$$\\$ (ii) The electrode reaction for 1 mol of FeO is$\\$ $FeO+\dfrac{1}{2}O_2\to \dfrac{1}{2}Fe_2O_3\\ i.e.,Fe^{2+}\to Fe^{3+}+e^-$ $\\$ $\therefore$ Quantity of electricity required =$1 F = 96500 C$

23   Give reason why a finely divided substance is more effective as an adsorbent.

##### Solution :

Quantity of electricity passed =$(5A) * (20 * 60 sec) = 6000 C$$\\ Ni^{2+}+2e^{-}\to Ni \\ Thus, 2F, i.e., 2 * 96500 C of charge deposit = 1 mole of Ni = 58.7 g\\ \therefore 6000 C of charge will deposit\\ =\dfrac{58.7*6000}{2*96500}=1.825 g \ \ of \ \ Ni Adsorption is a surface phenomenon. Therefore adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorptions increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent. Given: I = 1.5 A, W = 1.5 g of Ag_ t = ?, E = 108, n = 1 \\ Using Faraday’s 1 st law of electrolysis W = ZIt or W =\dfrac{E}{nF} It \\ or t=\dfrac{1.45*96500}{1.5*108}=863.73 seconds.\\ Now for Cu, W1 = 1.45 g of Ag E 1 = 108, W 2 = ? E 2 = 31.75\\ Form Faraday’s 2 nd law of electrolysis \dfrac{W_1}{W_2}=\dfrac{E_1}{E_2}$$\\$ $\dfrac{1.45}{W_2}=\dfrac{108}{31.75} \therefore W_2=\dfrac{1.45*31.75}{108}\\ =0.426 g \ of \ Cu$ $\\$ Similarly, for Zn, W 1 = 1.45 g of Ag, E 1 = 108, W2 = ? E2 = 32.65$\\$ Using formula, $\dfrac{W_1}{W_2}=\dfrac{E_1}{E_2}\\ \dfrac{1.45}{W_2}=\dfrac{108}{32.65}\\ \therefore W_2=\dfrac{1.45*32.65}{108}=0.438$ of Zn