Electro - Chemistry-8 · KnowledgeBin.org

HARD TYPE

Q.1. Standard potential of following cell is 0.23V at 15⁰C and 0.21V at 35⁰C
Pt H₂(g)|HCl(ag.)|AgCl(s)|Ag(s)
(i) Calculate

n⁰ &

S⁰ for cell rxn . by assuming that these quantities remains unchanged in range 15⁰ & 35⁰C.
(ii) Calculate solubility of AgCl in H₂O at 25⁰C. Given, standard reduction potential of Ag⁺(ag.)/Ag(s) couple is 0.8v at 25⁰C.

Ans: Pt H₂(g)|HCl(ag.)|AgCl(s)|Ag(s)

H₂

H⁺ + e^- (Anode)
AgCl + e^-

Ag + Cl^- (Cathode)

H₂ + AgCl

n⁺ + Ag + Cl^-
-

G⁰ = nE⁰F = 1 x 0.23 x 96500 = 2219 SJ (at 25⁰ c)
-

G⁰ = nE⁰F = 1 x 0.21 x 96500 = 2026 SJ (at 35⁰ c)

G⁰ =

H⁰ - T

S⁰
- 22195 =

H⁰ - 298 x

S⁰ ................................. (i)
- 20265 =

n⁰ - 308 x

S⁰ ................................. (ii)
On solving

S⁰ = -96.5 J &

H⁰ = 49.987 KJ

(ii) Ag

Ag⁺ + e^- E⁰_OP = - 0.8 V
Agcl + e^-

Ag + Cl^-

AgCl

Ag⁺ + Cl^-

E_cell =

log[AG⁺] +

At eqm,
E_cell = 0

log[Ag⁺][Cl^-] = 0.059 logK_{SP AgCl}
    - 0.8 + 0.22 = 0.059 logK_SP
      K_SP = 1.47 x 10^-10
      Solubility of AgCl =

= 1.21 x 10^-5 mol/L

Q.2. Overall formation constant for reaction of 6 mole of CN^- with cobalt (II) is 1 x 10¹⁹. Calculate formation constant for rxn. of 6 mole of CN^- with cobalt (III). Given that

CO

+ e^-

= - 0.83 V
CO³⁺ + e^-

CO²⁺

= 1.82 V

Ans: CO

+ e^-

= 0.83V
CO³⁺ + e^-

CO²⁺

= 1.82 V

+ CO³⁺

CO²⁺ + CO

E_cell = E⁰_cell -

log₁₀

Also, 6CN^- + CO²⁺

& 6CN^- + CO³⁺

E_cell = E⁰_cell +

log₁₀

At eqm. E_cell = 0
0 = 1.82 - (- 0.82) +

log₁₀

= 8.23 x 10⁴⁴

= 8.23 x 10⁶³

Q.3. Show that potential areadditive for process in which half reaction are added to yeild on overall rxn. but they arenot additive when added to yeild a third half rxn.

Ans: Case - I: When two half rxnx. are added to give an over all rxn., no. of moles of electrons involved in each half rxn. & over all rxn. are necessarily same.
M₁