Potentials = Ag=.799 Zn=-.763 D Standard Reduction (Elec x C Chegg Study Guided

Question

Potentials = Ag=.799 Zn=-.763 D Standard Reduction (Elec x C Chegg Study Guided Sx akeAssignment/takeCovalentActivity dot?ocator-assignment-take&takeAssignmentSessiontocators; assignment take Use the Refereaces to access important valoes if needed for this question When the Ag concentration is 5.06 104M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.363V. What is the Za2 concentration Answer Retry Entire Group 7 more group attenpts remaining 8 5 6

Explanation / Answer

     Zn(s) ------------------> Zn^2+ (aq) + 2e^-           E0 = 0.763v

     2Ag^+ (aq) + 2e^- --------> 2Ag(s)                    E0 = 0.799v

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    Zn(s) + 2Ag^+ (aq) -----------. Zn^2+ (aq) + 2Ag(s)   E0cell = 1.562v

     n = 2

[Ag^+]   = 5.06*10^-4 M

    Ecell   = 1.363v

Ecell     =   E0cell - 0.0592/n logQ

1.363      = 1.56 - 0.0592/2 log[Zn^2+]/[Ag^+]^2

1.363-1.562 = -0.0296log[Zn^2+]/(5.06*10^-4)^2

-0.199      = -0.0296log[Zn^2+]/(5.06*10^-4)^2

log[Zn^2+]/(5.06*10^-4)^2    = 0.199/0.0296

log[Zn^2+]/(5.06*10^-4)^2      = 6.723

[Zn^2+]/(5.06*10^-4)^2          = 10^6.723

[Zn^2+]/(5.06*10^-4)^2        = 5284452.5

[Zn^2+]                               = 5284452.5*(5.06*10^-4)^2

[Zn^2+]                                  = 1.353M >>>>>answer

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