Question

3. (2 pts.) For the following electrochemical cell at 25°C:

Zn(s) | Zn²⁺ (aq) || H⁺ (aq) | Pt,H₂ (g) (P=1.00 atm)

1. calculate DG° in kJ
2. calculate E_cell for [H⁺] = 0.8 M and [Zn²⁺] = 0.5 M
3. calculate K at 25°C

Answer 1

Zn(s) | Zn^{​​​​​ 2+} (aq) || H ⁺ (aq) | pt , H ₂ (g) ( P= 1.00 atm )

For Zn | Zn ²⁺ standard electrode potential is represent as

E⁰​​​​​ _{Zn | Zn^{​​​​​2+}} = 0.76 V and for hydrogen , E^{​​​​​​0} _{H2 | 2H​​​​​⁺} = 0.00 V

L. H .E. Zn (s) Zn​​​^{​​​​​2+} (aq)  + 2e  

R.H.E. 2H ⁺ (aq ) +2e   H_{​​​​​​2} (g)

Hence the cell reaction is  

Zn(s) + 2H⁺ (aq) Zn^{​​​​​2+} (aq) + H_{​​​​​​2} (g) ............ (1)

Hence the two electron take part in this reaction . Hence n= 2 ,

Hence the total E ⁰ cell is = E⁰​​​​​ _{Zn | Zn^{​​​​​2+}} + E^{​​​​​​0} _2H+ _{| H2}

_{=( 0.76 + 0.0 ) V}

_{= 0.76 V}

a) We know that ∆G⁰ = - nFE^{​​​​0}

We know that F = 96500C

    ∆G⁰ = -( 2 mol × 96500 C mol^{​​​​-1} × 0.76 V )

∆G⁰ = - 146680 CV = -146680 J

∆G⁰ = -146.68 KJ

b) From nernst equation we get (using equation 1)

For solid [Zn] = 1

  

Here, For solid [Zn] = 1 [Zn^{​​​​​2+}] = 0.5 M

[H⁺] = 0.8 M  

E = 0.756 V  

Hence the E _cell = 0.756 V

C) At equilibrium E_{​​​​​​cell} = 0

E^{​​​​​​0}_{​​​Cell} = (0.059/n) logK

0.76 = (0.059/2) logK

logK = 25.7627

K = 5.79 × 10²⁵

Hence K = 5.79 × 10²⁵

Here T=25⁰ C = (25+273) K = 298 K

R = 8.314 J mol^{​​​​-1} K^{​​​​​​-1}