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Chemistry Answer the following in brief.

Electrochemistry · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 11 questions.

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Question 13 Marks
Resistance of conductivity cell filled with $0.1 M \ KCl$ solution is $100 \ ohms,$ calculate the conductivity and molar conductivity of $0.02 M \ KCl$ solution. [Given: Conductivity of $0.1 M \ KCI$ solution is $1.29 \ Sm^{-1}.]$
Answer
Given that:
Concentration of the $KCl$ solution =$ 0.1 \ mol \ L^{−1}$
Resistance of cell filled with $0.1 \ mol \ L^{−1} KCl$ solution $= 100 \ ohm$
Cell constant $= G^* =$ conductivity $\times$ resistance
$1.29\times 10−2 \ ohm−1\ cm−1 \times 100 \ ohm = 1.29 \ cm−1 = 129 m−1$
Cell constant for a particular conductivity cell is a consant.
Conductivity of $0.02 \ mol \ L ^{-1} KCl$ solution $=\frac{\text { Cell constant }}{\text { Resistance }}$
$\frac{G *}{R}=\frac{129 \ m ^{-1}}{520 \ ohm }=0.248 \ Sm ^{-1}$
Concentration$ = 0.02 \ mol \ L^{−1}$
$= 1000 \times 0.02 \ mol \ m^{−3} = 20 \ mol \ m^{−3}$
Now,Molar conductivity $=$
$\Lambda_m=\frac{\kappa}{c}=\frac{248 \times 10^{-3} \ Sm ^{-1}}{20 \ mol \ m ^{-3}}=124 \times 10^{-4} \ Sm ^2 \ mol ^{-1}$
Therefore, the molar conductivity of $0.02 \ mol \ L^{−1} \ KCl$ solution is
$124 \times 10^{-4} \ Sm ^2 \ mol ^{-1}$
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Question 33 Marks
Define terms: (a) Electrochemical series (b) Corrosion. Write two applications of electrochemical series.
Answer
electrochemical series:
EMF series is defined as the arrangement of the electrode with the electrode half-reaction in order of decreasing standard potentials.
Corrosion:
Corrosion is defined as a process where materials, usually metals, deteriorate as a result of a chemical reaction with air, moisture, chemicals, etc.
For example, iron, in the presence of moisture, reacts with oxygen to form hydrated iron oxide.
Applications of electrochemical series:
The relative strength of reducing agents:
1. The species on the right side of half-reactions are reducing agents.
2. The species appearing at the bottom right side of half-reactions associated with large negative E° values are the effective electron donors. They serve as strong reducing agents.
3. The strength of reducing agents increases from top to bottom as E° values decrease.
Spontaneity of redox reactions:
1. Spontaneity of redox reactions can be predicted from values of electrode potential in electrochemical series.
2. If the species with a higher E° value is reduced (accepts electrons) and that with a lower E° value is oxidized (donates electrons), then the redox reaction in a galvanic cells is spontaneous.
3. The standard cell potential must be positive for a cell reaction to be spontaneous under the standard conditions.
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Question 43 Marks
Define electrochemical series. Write its applications.
Answer
electrochemical series:
$\text{EMF}$ series is defined as the arrangement of the electrode with the electrode half$-$reaction in order of decreasing standard potentials.
electrochemical series of applications:
$1.$ Oxidizing and Reducing Strengths: The electrochemical series helps to identify the substances that are good oxidizing agents and reducing agents. All the substances appearing on the top of the series behave as good reducing agents. All the substances appearing at the bottom of the table are good oxidizing agents.
$2.$ Displacement reactions: A metal higher in the series will displace the metal from its solution which is lower in the series. A metal higher in the series has a greater tendency to provide electrons to the cations of the metal to be precipitated. The metal having low standard reduction potential will displace the metal from its salt's solution which has a higher value of standard reduction potential.
$3.$ Predicting the Liberation of Hydrogen Gas from Acids by Metals: All metals having negative electrode potentials $(- E^\circ )$ show a greater tendency of losing electrons as compared to hydrogen. So, when such a metal is placed in an acid solution, the metal gets oxidized, and $H^+$ ions get reduced to form hydrogen gas. Thus, the metals having $- E^\circ$ values liberate hydrogen from acids.
$4.$ Predicting the Feasibility of a Redox Reaction: Depending on the $E^\circ$ values of the two electrodes feasibility of the given redox reaction can be found out. A redox reaction is feasible only if the species which has higher potential is reduced i.e., accepts the electrons and the species which has lower reduction potential is oxidized i.e., loses electrons.
$5.$ Calculation of the EMF of the Cell: If the $\text{EMF}$ of the cell is positive, the reaction is feasible in the given direction and the cell is correctly represented. If it is negative, the cell reaction is not feasible in the given direction and the cell is wrongly represented.
$6.$​​​​​​​ Comparison of Reactivities of Metals: The relative ease with which the various species of metals and ions may be oxidized or reduced is indicated by the reduction of potential values. The metals with lower reduction potential are not reduced easily but are easily oxidized to their ions losing electrons.
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Question 53 Marks
How many faradays of electricity are required to produce $13 g$ of aluminium from aluminium chloride solution? $($Given: molar mass of $Al = 27.0 \ g \ mol^{-1})$
Answer
$AlCl _3 \longrightarrow Al ^{+3}+3 Cl ^{-}$
$Al ^{+3}+3 e ^{-} \longrightarrow Al$
$1 \ mole$ of $Al$ requires passage of $3 \ moles$ of electrons. The charge on $3 \ moles$ of $e^-$ is $3$ Faraday.
Moles of $Al$ produced $=\frac{\text { mass of } Al }{\text { molar mass of } Al }$
$=\frac{13}{27}$
$= 0.48 \ moles$
As $3F$ of electricity produces $1 \ mole$ of $Al$
$\therefore$ No. of faradays of electricity required to produce $0.48 \ mole$ of $Al$
$= 0.48 \times 3$
$= 1.44$ Faraday
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Question 63 Marks
$0.05 M \ NaOH$ solution offered a resistance of $31.6 W$ in a conductivity cell at $298 K$. If the cell constant of the cell is $0.367 \ cm^{-1}.$ Calculate the molar conductivity of $NaOH$ solution.
Answer
Given:
Cell constant $(b) = 0.367 \ cm^{–1}$
Molar concentration $(C) = 0.05 M \ NaOH$
Resistance $(R) = 31.6 \Omega$
To find: Molar conductivity 
Solution:
$∧_{m(NaOH)} =$ ?
$k_{( NaOH )}=\frac{b}{R_{( NaOH )}}=0.01161 \Omega^{-1} \ cm ^{-1}$
$\wedge_{m(N a O H)}=\frac{k \times 1000}{C}$
$=\frac{0.01161 \times 1000}{0.05}$
$= 232.2 \ \Omega ^{-1} \ cm^2 \ mol^{-1}$
Molar conductivity $= ∧_m = 232.2 \ \Omega ^{-1} \ cm^2 \ mol^{-1}$
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Question 73 Marks
Conductivity of a solution is $6.23 \times 10^{-5} \Omega^{-1} cm ^{-1}$ and its resistance is $13710 \Omega$. If the electrodes are $0.7 cm$ apart, calculate the cross-sectional area of electrode.
Answer
ross-sectional area of electrode:
$k=\frac{1}{R} \cdot \frac{l}{a}$
$a=\frac{1}{R} \cdot \frac{l}{k}$
$a=\frac{1}{13710} \cdot \frac{0.7}{6.23 \cdot 10^{-5}}$
$a=0.82 cm ^2$
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Question 93 Marks
Calculate $E _{\text {cell }}$ and $\Delta G$ for the following at $28^{\circ} C$ :
$Mg _{( s )}+ Sn ^{2+}(0.04 M ) \longrightarrow Mg ^{2+}(0.06 M )+ Sn _{( s )} E_{\text {cell }}^0=2.23 V$
Is the reaction spontaneous?
Answer
$M g_s+S n_{a q}^{2+} \rightarrow M g_{a q}^{2+}+S n_s$
$\left[ Sn ^{2+}\right]=0.04 M\left[M g^{2+}\right]=0.06 M$
$Q =\frac{M g^{2+}}{S n^{2+}}=\frac{0.06}{0.04}=1.5$
$E_{\text {cell }}=2.23 V$
$R = 8.314 \ JK^{-1} \ mol^{-1}$
$T = 28^\circ C = 28 + 273 = 301K$
$n = 2$
$F = 96500 \ C/mol e^-$
$\therefore E_{\text {cell }}=E_{\text {cell }}-\frac{2.303 R T}{n F} \log _{10} Q$
$=2.23-\frac{2.303 \times 8.314 \times 301}{2 \times 96500} \log _{10}(1.5)$
$=2.23-\frac{2.303 \times 8.314 \times 301}{2 \times 96500} \times(0.1760)$
$= 2.23 - 0.005255$
$E_{\text {cell }}=-2.2247 V$
$\Delta G=n F E_{\text {cell }}$
$\Delta G=-2 \times 96500 \times 2.2247$
$\Delta G=-429367 J$
$\because \triangle G$ is negative therefore reaction is spontaneous.
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Question 103 Marks
How much quantity of electricity in coulomb is required to deposit $1.346 \times 10^{-3} \ kg$ of $Ag$ in $3.5$ minutes from $AgNO _3$ solution? $($Given: Molar mass of $Ag$ is $108 \times 10^{-3} \ kg \ mol ^{-1})$
Answer
Given:
Mass of $Ag$ deposited $= 1.346 \times 10^{-3} \ kg$
Time$ (t) = 3.5 \ min = 3.5 \times 60s$
Molar mass of $Ag = 108 \times 10^{-3} \ kg \ mol^{-1}$
To find:
Quantity of electricity required in coulomb:
Formulae:
$1.$ Mole ratio $=\frac{\text { Moles of product formed in half reaction }}{\text { Moles of electrons required in half reaction }}$
$2.$ Mass of the substance produced $=$
$\frac{ I ( A ) \times t ( s )}{96500( C / mol \text { e- }) \times \text { Mole ratio } \times \text { Molar mass of substance }}$
$3.$ Quantity of electricity in coulomb $(Q) = I ($ in amp $) \times t($ in sec$)$
Calculation:
The half reaction for the formation of $Ag$ is,
$\left.\operatorname{Ag}_{( aq }\right)^{+}+ e ^{-} \longrightarrow \operatorname{Ag}_{( s )}$
From formula$ (1),$
Mole ratio $=\frac{\text { Moles of } Ag }{\text { Moles of electrons }}$
$=\frac{1( mol \ Ag )}{1\left( mole ^{-}\right)}$
$= 1 \ mol  \ Ag/mol e^-$
From formula $(2),$
$1.346 \times 10^{-3}$
$=\frac{I \times 3.5 \times 60}{96500} \times 1 \times 108 \times 10^{-3}$
$\therefore I=\frac{1.346 \times 10-3 \times 96500}{3.5 \times 60 \times 108 \times 10^{-3}}$
$=\frac{129889 \times 10^{-3}}{22680 \times 10^{-3}}$
$= 5.727 A$
From formula $(3),$
$Q = It = 5.727 \times 3.5 \times 60 = 1202.67 C$
$\therefore$ The quantity of electricity required is $1202.67 C.$​​​​​​​
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Question 113 Marks
Write the cell representation and calculate equilibrium constant for the following redox reaction:
$Ni _{( s )}+2 Ag _{( aq )}^{+}(1 M ) \longrightarrow Ni ^{2+}{ }_{\text {(aq) }}(1 M )+2 Ag _{\text {(s) }}$ at $25^{\circ} C ; E _{ Ni }^0=-0.25 V$ and $E _{ Ag }^0=0.799 V$
Answer
$N i_{(s)}+2 A g_{(a q)}^{+}(1 M) \rightarrow N i_{(a q)}^{2+}(1 M)+2 A g_{(s)}$ at $25^{\circ} C$
$E_{n i}^{+}=-0.25 V$ and $E_{A g}^{+}=0.799 V$
Cell representation :
$N i_{(s)}\left|N i_{(a q)}^{2+}(1 M)\right|\left|A g_{a q}^{+}(1 M)\right| A g(s)$
Calculation of equilibrium constant
$E_{\text {cell }}^{\circ}=\frac{0.0592}{n} \log _{10} K$
$E_{\text {cell }}^{\circ}=E_{ Ag }^{\circ}-E_{ Ni }^{\circ}$
=0.799-(-0.25)=1.049 V
Hence $1.049=\frac{0.0592}{2} \log _{10} K$
$\log _{10} K=\frac{1.049 \times 2}{0.0592}=35.44$
$K=$ anti $\log (35.44)$
$=2.754 \times 10^{35}$
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