UNIT 9: Electro Chemistry - Online Test

1F = 96500 C = 1 mole of e⁻ = 6.023 ×10²³ e⁻

∴ 9650 C = [ 6.22 ×10²³ / 96500 ] × 9650 = 6.022×10²²

= 6.022×10²²

Mn²⁺ + 2e⁻ → Mn (Eº_red) = -1.18V

2[Mn²⁺ → Mn³⁺ + e^- ](Eº_ox) = −1.51V

3Mn²⁺ → Mn + 2Mn³⁺        Eº_cell?

Eº_cell = ( Eº_ox )+ ( Eº_red )

= −1.51 −1.18 and non spontaneous

= −2.69V

Since E^o is –ve ∆G is +ve and the given forward cell reaction is non – spontaneous.

Anodic oxidation: (Reverse the given reaction)

( E^o_ox ) = 0.76V cathodic reduction

∴ E_cell = ( E^o_ox )+ ( E_red^o )

 = 0.76 + 0.34 = 1.1V

= 1.1V

Λ = κ/M ×10⁻³ mol⁻¹ m³

= [( 5.76 × 10⁻³ S cm⁻¹ ×10⁻³ ) / (0. 5)]  mol⁻¹m³

= [(5.76 × 10⁻³ × 10⁻³ ×10⁶ ) / (0.5)] S cm⁻¹mol⁻¹ cm³ .

= 11.52 S cm² mol⁻¹

( Λ^º )_HoAC = ( Λ^º)_HCl + ( Λ^º)_NaOAC - ( Λ^º)_NaCl

= (426.2 + 91) −(126.5)

= 390.7

1F = 96500 C = charge of 1 mole of e^- =

charge of 6.022 ×10²³ e⁻

 7MnO^-₄ + 5e⁻ → Mn²⁺ + 4H₂O

5 moles of electrons i.e., 5F charge is required.

 m=ZIt

(41min 40sec = 2500 seconds)

Z = m/ (n × 96500) = 40 / (2 × 96500)

m = Zit

 = ( 40 × 3.86 × 2500 ) / (2 × 96500)

= 2g

m=ZIt

(mass of 1 mole of Cl₂ gas = 71)

t = m/ ZI  (∴mass of 0.1mole of Cl₂ gas = 7.1 g mol⁻¹ )

 = 7.1 / [71/(2 ×96500) × 3 ]

 (2 Cl- → Cl₂ +2e^- )

= (2 × 96500 × 7.1) / (71× 3)

 = 6433.33sec

 = 107.2 min

 Q =It

 = 1A×60S

96500 C charge ≡ 6.022 ×10²³ electrons

60 C charge ≡ [ 6.022 ×10²³ /  96500  ] × 60

 = 3.744 ×10²⁰ electrons