Question

C152-02 Class Practice Chapter 21 Practice 4 Calculate the reduction potential of the Cu2+/Cu electrode when Cu2+1 -1.0 x10 the table of standard reduction potentials and the Nernst Equation) M. (hint: use A cell is constructed by immersing a strip of silver in 0.10 M AgNO, solution and a strip of lead in 1.0 M Pb(NO3), solution. A wire and salt bridge complete the cell. What is the potential for the cell? Chint: use the Nernst Equation to determine the reduction potentials of the two half.cells) Calate for the reaction below when [7n'] 1.00 M. H) - 1.00x10-6 M. and P. 100 atm. Zn(s) +2H2n+H.(g) (hint: use the Nernst Equation, E-E-0.0591 Log In

Answer 1

Cu2+ ( ceq) + 20 - Cu(s) E = € - 0.059 log & E = 0.14 rooss log Turn (here, na no. of è transferred] Ca27/6 2034 E = 0.34 -0.059 E = 0.34 -0.org lo = 0.34 - o.org x 8 = 0.104 = 0.104 volt -- 0.8 volt -0.13 volt electro Chemical Series. from Reaction : 2 Ag+ (aq) + Pb(s) Pb2+ (aq) + 2AGS) E cal - E Cathode - Canoce - 0.8 - (-0.13) = 0.93 vat.

E call = E cu rooss log & = 093-dogging Cleaning = 0.93 - Dogo log 10.12 -0.93 -0.org xx 0.871 volt Ez2+/zn -0.76 vort , Entln = 0 Reacton: Zn (s) + 24+ (aq) → Zn2+ (aq) + H2(g) E au = o Cathode - Eanode = o- 6076) -0.76 volt E = 60 - o.org) lug [Zn2+] the [nt]2 - 0.76- 09059) log 1x1 = 0.76 0.0751 x 8 - 0.523 volt