Class 12 Chemistry

# Electrochemistry

## NCERT Solution

Part 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, Hg^{2+}/Hg = 0.79 V, Mg^{2+}/Mg = -2.37 V, Cr^{3+}/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 Δ_{TGΘ 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)|Mg^{2+} (0.001 M)|| Cu^{2+} (0.001 M)|Cu(s)

**Answer:** Cell Reaction

`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)|Fe^{2+}(0.001 M)||H^{+}(1 M)|H_{2}(g)(1 bar)|Pt(s)

**Answer:** Cell reaction:

`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)|Sn^{2+}(0.050 M)|| H^{+}(0.020 M)|H_{2}(g) (1 bar)|Pt(s)

**Answer:** Cell reaction:

`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)|Br_{2}(l)||H^{+}(0.030 M)|H_{2}(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 Δ_{T}G^{Θ} and E^{Θ} for the reaction.

**Answer:** Here, Zn is oxidized and Ag_{2}O is reduced

E_{cell}°=E_{Ag2O|Ag}° - E_{Zn|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 λ_{m} is S m^{2 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 cm^{2} 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) | 0.001 | 0.010 | 0.020 | 0.050 | 0.100 |
---|---|---|---|---|---|

10^{2} `xx` κ/S m^{-1} | 12.37 | 11.85 | 23.15 | 55.53 | 106.74 |

Calculate Λ_{m} for all concentrations and draw a plot between Λ_{m} and C^{1/2. Find the value of Λm&Deg;.}

**Answer:**

Concentration(M) | 0.001 | 0.010 | 0.020 | 0.050 | 0.100 |
---|---|---|---|---|---|

10^{2} `xx` κ/S m^{-1} | 12.37 | 11.85 | 23.15 | 55.53 | 106.74 |

Λ_{m} | 123.7 | 118.5 | 115.8 | 111.1 | 106.7 |

`c^(1/2)(M^(1/2))` | 0.0316 | 0.100 | 0.141 | 0.224 | 0.316 |

Here, Λ° = 124.0 S cm^{2} mol^{-1}

Daniell Cell

InText Solution

NCERT Solution 1

NCERT Solution 2