Class 12 Chemistry

# Electrochemistry

## NCERT SolutionPart 1

Question 1: Arrange the following metals in the order in which they displace each other from the solution of their salts. (Al, Cu, Fe, Mg and Zn)

Answer: Mg, Al, Zn, Fe, Cu, Ag

Question 2: Given the standard electrode potentials.

K+/K = -2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = -2.37 V, Cr3+/Cr = -0.74 V

Arrange these metals in their increasing order of reducing power.

Answer: We know that reducing power of a metal increases with increase in oxidation potential. So, correct order is Ag > Hg > Cr > Mg > K

Question 3: Depict the galvanic cell in which the reaction Zn(s)+2Ag^(+)(aq)→Zn^(2+)(aq)+2Ag(s) takes place. Further show

• Which of the electrode is negatively charged?
• The carriers of the current in the cell.
• Individual reaction at each electrode.

Answer: The cell is represented as follows:

Zn(s)|Zn^(2+)(aq)||Ag^(+)(aq)|Ag(s)

Anode (negatively charged electrode): Zinc electrode

Current will flow from silver to copper in the external circuit.

Reactions at each electrode are as follows:

Anode: Zn(s)→Zn^(2+)(aq)+2e^(-)

Cathode: 2Ag^(+)(aq)+2e^(-)→2Ag(s)

Question 4: Calculate the standard cell potentials of galvanic cell in which the following reaction takes place:

2Cr(s)+3Cd^(2+)(aq)→2Cr^(3+)(aq)+3Cd

Fe^(2+)(aq)+Ag^(+)(aq)→Fe^(3+)(aq)+Ag(s)

Calculate the ΔTGΘ and equilibrium constant of the reactions.

Answer: E_(ce\ll)^Θ=E_(ca\th\od\e)^Θ-E_(an\od\e)^Θ

=-0.40V-(-0.74V)=0.34V

Δ_T\G°=-nF\E_(ce\ll)^Θ

=-6xx96500 C mo\l^(-1)xx0.34 V

=-196860 C V mol-1=-196.86 kJ mol-1

Δ_TG°=-2.303 RT log K

Or, 196860=2.303xx8.314xx298 log K

Or, log K = 34.5014

Or, K = Antilog 34.5014=3.172xx10^(34)

E_(ce\ll)^Θ=0.80V-0.77V=0.03 V

Δ_T\G°=-nF\E_(ce\ll)^Θ

=-1xx96500xx0.03

=-2895 J mo\l^(-1)=-2.895 kJ mol-1

Δ_TG°=-2.303 RT log K

Or, -2895=-2.303xx8.314xx298xx log K

Or, log K= 0.5074

Or, K = Antilog (0.5074) = 3.22

Question 5: Write the Nernst equation and emf of the following cells at 298 K.

Mg(s)|Mg2+ (0.001 M)|| Cu2+ (0.001 M)|Cu(s)

Mg+Cu^(2+)→Mg^(2+)+Cu(n=2)

Nernst equation:

E_(ce\ll)=E_(ce\ll)^Θ-(0.0591)/(2)lo\g\([Mg^(2+)])/([Cu^(2+)])

=0.34-(-2.37)-(0.0591)/(2)lo\g(10^(-3))/(10^(-4))

=2.71-0.02955=2.68 V

Fe(s)|Fe2+(0.001 M)||H+(1 M)|H2(g)(1 bar)|Pt(s)

Fe+2H^(+)→Fe^(2+)+H_2(n=2)

Nernst equation:

E_(ce\ll)=E_(ce\ll)^Θ-(0.0591)/(2)lo\g\([Fe^(2+)])/([H^(+)]^2)

=0-(-0.44)-(0.0591)/(2)lo\g\(10^(-3))/((1)^2)

=0.44-(0.0591)/2xx(-3)

=0.44+0.0887=5.287 V

Sn(s)|Sn2+(0.050 M)|| H+(0.020 M)|H2(g) (1 bar)|Pt(s)

Sn+2H^(+)→Sn^(2+)+H_2(n=2)

Nernst equation:

E_(ce\ll)=E_(ce\ll)^Θ-(0.0591)/(2)lo\g\([Sn^(2+)])/([H^(+)]^2)

=0-(-0.14)-(0.0591)/(2)lo\g\(0.05)/(0.02)^2

=0.14-(0.0591)/(2)lo\g\125

=0.14-(0.0591)/(2)(2.0969)=0.075 V

Pt(s)|Br-(0.10 M)|Br2(l)||H+(0.030 M)|H2(g) (1 bar)|Pt(s)

Answer: cell reaction: 2Br^(-)+2H^(+)&Br_2+H_2(n=2)

E_(ce\ll)=E_(ce\ll)^Θ-(0.0591)/(2)lo\g\(1)/([Br^(-)]^2[H^(+)]^2)

=0-1.08-(0.00591)/(2)lo\g\(1)/((0.01)^2(0.03)^2)

=-1.08-(0.0591)/(2)lo\g(1.111xx10^7)

=-1.08-(0.0591)/(2)xx7.0457

=-1.08-0.208=-1.288 V

Question 6: In the button cells widely used in watches and other devices the following reaction takes place.

Zn(s)+Ag_2O(s)+H_2O(l)→Zn^(2+)(aq)+2Ag(s)+2OH^(-)(aq)

Determine ΔTGΘ and EΘ for the reaction.

Answer: Here, Zn is oxidized and Ag2O is reduced

Ecell°=EAg2O|Ag° - EZn|Zn2+°

=0.344+0.76=1.104 V

ΔG=-nF\E_(ce\ll)

=-2xx96500xx1.104 J

=-2.13xx10^5 J

Question 7: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Answer: Conductivity: Conductance of a solution of 1 cm length and having 1 sq cm as area of cross-section is called conductivity of the solution.

Molar Conductivity: Molar conductivity is the conductivity of the solution divided by molar concentration of the solution. It is given by following equation.

λ_m=κ/c

If the unit of conductivity is S m-1 and the unit of concentration is mol m-3 then unit of λm is S m2 mol-1

Variation of Conductivity and Molar Conductivity with Concentration

Conductivity always decreases with decrease in concentration of both, weak and strong electrolytes. This happens because the number of ions per unit volume decreases on dilution.

Molar conductivity of a solution increases with a decrease in concentration. This happens because the total volume V of solution containing one mole of electrolyte also increases. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

Question 8: The conductivity of 0.20 M solution of KCl at 298 K is 0.248 S cm-1. Calculate it molar conductivity.

Answer: Λ_m=(κxx1000)/text(Molarity)

=(0.0248 S cm^(-1)xx1000 cm^3L^(-1))/(0.20 mo\l L^(-1))

=124 S cm2 mol-1

Question 9: The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 &Ohm;. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 xx 10-3 S cm-1.

Answer: Cell constant = text(Conductivity)/text(Conductance)

= Conductivity xx Resistance

=0.146xx10^(-3) S cm^(-1)xx1500 &Ohm;

=0.219 cm-1

Question 10: The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

 Concentration(M) 102 xx κ/S m-1 0.001 0.01 0.02 0.05 0.1 12.37 11.85 23.15 55.53 106.74

Calculate Λm for all concentrations and draw a plot between Λm and C1/2. Find the value of Λm&Deg;.

 Concentration(M) 102 xx κ/S m-1 Λm c^(1/2)(M^(1/2)) 0.001 0.01 0.02 0.05 0.1 12.37 11.85 23.15 55.53 106.74 123.7 118.5 115.8 111.1 106.7 0.0316 0.1 0.141 0.224 0.316

Here, Λ° = 124.0 S cm2 mol-1

Daniell Cell

InText Solution

NCERT Solution 1

NCERT Solution 2