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$$\\$ Applying Nernst equation,$\\$ $E_{H^-\dfrac{1}{2}H_2} = E^o_{H^\dfrac{1}{2}H_2}-\dfrac{0.0591}{n}log\dfrac{1}{[H^+]}$$\\$ $=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$$\\$