Electrochemistry

Chemistry

NCERT

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

Solution :

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