# 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,$c(1-\alpha )\qquad c\alpha \qquad c\alpha$ $\\$ $\therefore K_a=\dfrac{c\alpha.c\alpha}{c(1-\alpha)}=\dfrac{c\alpha^2}{1-\alpha}\\ =\dfrac{0.025*(0.114)^2}{1-0.114}=3.67*10^{-4}$