Question

Write the Nernst equation and EMF of the following cells at 298K: a. Mg(s)|Mg2+(0.001M)||Cu2+(0.0001M)|Cu(s) b. Fe(s)|Fe2+(0.001M)|H⊕(1M)|H2(g)(1bar)|Pt(s) c. Sn(s)|Sn2+(0.050M)||H⊕(0.020M)|H2(g)(1bar)∣Pt(s) d. Pt(s)|Br2(1)|Brc-(0.010M)||H⊕(0.030M)|H2(g)(1bar)|Pt(s) Select the correct answer from above options

Answer 1

Correct Answer - a.2.68V , b.0.523V , c.0.078V , d.-1.288V a. Cell reaction :Mg+Cu2+→mg2+=Cu(n=2) Ecell=E c- cell - 0.0591 2 log [Mg2+] [Cu2+] =0.34-(-2.37)- 0.591 2 log 10-3 10-4 =2.71-0.02955=2.68 b. Cell reaction Fe+2H⊕→Fe2++H2(n=2) Ecell=E c- cell - 0.0591 2 log([Fe2+]) [H⊕] =0-(-0.44)- 0.0591 2 log 10-3 (1)2 =0.44+0.0887=0.523V c.Cellreaction: Sn+2H^(o+) rarr Sn^(2+)+H_(2) (n=2) E_("cell")=E_("cell")^(c-)-(0.0591)/(2)log ([Sn^(2+)])/([H^(o+)]) =0-(-0.14)-(0.0591)/(2)log (0.05)/((0.02)^(2)) =0.14-(0.0591)/(2)log125 =0.14-(0.0591)/(2)(2.0969)=0.078V<br.d.Cellreaction :2Br^(c-)+2H^(o+) rarr Br_(2)+H_(2) E_("cell")=E_(cell)^(c-)-(0.0591)/(2)log (1)/([Br^(c-)][H^(o+)]) =(0-1.08)-(0.591)/(2)log (1)/((0.01)^(2)(0.03)^(2)) =-1.08-(0.0591)/(2)log(1.111xx10^(7)) =-1.08-(0.0591)/(2)(7.0457) =-1.08-0.208=-1.288VThus,⊗tionwillouratthehydro≥ne≤ctrodeandreductiononthe Br_(2)e≤ctrode. e_("cell")=1.288V`