Question

Write the Nernst equation and em! of the following cells at 298 K: 

(i) Mg(s)|Mg⁺²(0.001M) ||Cu⁺²(0.0001M)|Cu(s) . 

(ii) Fe(s)|Fe⁺²(0.001 M)||H⁺(1M)H₂(g) (1bar)|Pt(s) 

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

(iv) Pt(s)|Br₂(l)|Br^-(0.010 M)||H⁺(0.030 M)| H₂(g) (1 bar)|Pt(s).

Answer 1

(i) E^θCr³⁺ /Cr = 0.74V 

E^θCd²⁺ / Cd = - 0.40V 

The galvanic cell of the reaction IC depicted as : 

Cr(s)|Cr³⁺ (aq) || (Cd²⁺ (aq) Cd(s) 

Now, the standard cell potential is 

E^θ_cell = E^θ_R - E^θL 

= -40 - (-0.74) = +0.34V 

∆_rG^θ = -nFE^θ_cell 

In the given equation, n = 6 

F = 96487 C mol^-1 

E^θ_cell = + 0.34V 

Then, ∆_rG^θ = -6 × 96487 mol^-1 × 0.34V 

= -196833.48 CV mol^-1 

= -196833.48 J mol^-1 

= -196.83 KJ mol^-1 

Again ∆_rG^θ = -RT In K

=> ∆_rG^θ = -2.303RT log K

(ii) E^θ Fe³⁺ /Fe²⁺ = 0.77V 

E^θ Ag^- /Ag = 0.80V 

For the given reaction, the Nernst equation can be given as:

(iii) Far the givën reaction, the Nernst equation can be given as:

0.14 – 0.0295 × log 125 

= 0.14-0.062 

= 0.078 V 

= 0.08 V (approx)

(iv) For the given reaction, the nernst equation can be given as:

= - 1.09 - 0.02955 × log ( 1.11× 10⁷) 

= - 1.09 - 0.02955(0.0453 + 7) 

= -1.09 - 0.208 = -1.298 V.