question_answer 20)

Write the Nernst equation and rind e.m.f. of the following cells at 298 K. Mg(s)/Mg2+ (0.001 M) || Cu2+s(0.0001 M) | Cu(s) E0Mg2+/Mg = ? 2.37 V, E0Cu2+ /Cu = 0.34 V (ii) Fe (s)/Fe2+ (0.001 M) || H+ (1M)/H2 (g) (1 bar) | Pt (s) E0Fe+2/Fe = 0.44 V (iii) Sn (s)/Sn2+ (0.050 M) || H+ (0.020 M) | H2 (g) (1 bar) I Pt (s) (iv) Pt (s)/Br2 (1)/Br?(0.010 M) || H+ (0.030 M) |H2 (g) (1 bar) I Pt (s)  

Answer:

(i) The Nernst equation for this cell is                    ?. (i) Where = 0.34 ? (? 2.37) = 0.34 + 2.37 = 2.71 V Given: [Mg2+] = 0.001 M and [Cu2+] = 0.0001 M Put in equation (i), we get = 2.71 ? 0.0295 log 10 = 2.71 ? 0.0295 × 1 = 2.6805V. (ii)            ?. (i) = 0.00 V ? (?0.44V) = + 0.44 V [Fe2+] = 0.001 M and [H+] = 1M Put in the eq. (i), we get = 0.44 V ? 0.0295 [log 10?3] = 0.44 V ? 0.0295 × ? 3 log 10 = 0.44 V ? 0.0885 log 10 = 0.44 + 0.0885 V × 1           = 0.5285 V (iii) Cell reaction: Nernst equation: = 0.078 V. (iv) Cell reaction: Nernst equation: = 1.08 ? 0.208 = ?1.288V. Thus, oxidation will occur at the hydrogen electrode and reduction on the Br2 electrode. Ecell = 1.288 V.