This is one of my homework problems from my chemistry text, Ihave solved part A

Question

This is one of my homework problems from my chemistry text, Ihave solved part A and B but part C seems to be a dousy for me,please help if you can...here is the problem: Consider the voltaic cell 2 Ag+ (aq) + Cd (s) ------> 2 Ag (s) + Cd2+(aq) Operating (a) What is the Eocell for this cell? (b) If [Cd2+] = 2.0M and [Ag+] = 0.25M,what is Ecell? (c) If Ecell = 1.25V and [Cd2+] = .100M, what is[Ag+]? Useful info for problem: Ag+(aq) + e- ------->Ag(s)     Eo (V) = +0.7994 Cd2+(aq) + 2e- -------> Cd(s) Eo (V) = -0.403 This is one of my homework problems from my chemistry text, Ihave solved part A and B but part C seems to be a dousy for me,please help if you can...here is the problem: Consider the voltaic cell 2 Ag+ (aq) + Cd (s) ------> 2 Ag (s) + Cd2+(aq) Operating (a) What is the Eocell for this cell? (b) If [Cd2+] = 2.0M and [Ag+] = 0.25M,what is Ecell? (c) If Ecell = 1.25V and [Cd2+] = .100M, what is[Ag+]? Useful info for problem: Ag+(aq) + e- ------->Ag(s)     Eo (V) = +0.7994 Cd2+(aq) + 2e- -------> Cd(s) Eo (V) = -0.403

Explanation / Answer

a. The two half reactions are :                           Reaction at cathode :  Ag+(aq)+ e- -------> Ag(s)    Eo (V) = +0.7994 Reaction at anode :  Cd2+(aq) + 2e- -------> Cd(s) Eo (V) = -0.403 Overall reaction :      2 Ag+ (aq) +Cd (s) ------> 2 Ag (s) + Cd2+ (aq)   E0cell = E0 cathode -E0anode                                                                                                                  E0cell= +0.7994V - (-0.403V) = 1.2024V b.   Ecell = E0cell -(0.0592V/n)log{[Cd2+ (aq) ]/[Ag+(aq]2}                =1.2024V - (0.0592/2)log{(2.0)/(0.25)2}                =1.1578V c. Given data : Ecell = 1.25V and[Cd2+] = .100M, [Ag+]=?                         Ecell = E0cell -(0.0592V/n)log{[Cd2+ (aq) ]/[Ag+(aq]2}             1.25V = 1.2024V -(0.0592V/2)log{0.100/[Ag+ (aq]2}                  1.25V -  1.2024V = -(0.0592V/2)log{0.100/[Ag+(aq]2}                 0.0476V =  -(0.0592V/2)log{0.100/[Ag+(aq]2}                 (0.0476V*2)/ -0.0592V = log0.100 - 2log[Ag+ (aq]                    1.6081 = -1 -2log[Ag+ (aq]                    log[Ag+ (aq] = (1.6081 +1)/-2                                         =-1.3040                      [Ag+ (aq] = 10-1.3040                           [Ag+(aq] = 0.0497M                           [Ag+(aq] = 0.0497M

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