THE ANSWER is 0.531 Idk how did I end up in a value of 0.5398895 very close but

Question

THE ANSWER is 0.531

Idk how did I end up in a value of 0.5398895 very close but not right?

(8) 11. Copper and silver are in contact with a solution at 25°c containing cu? and Ag ions setting up a corrosion cell. 22 a Show the anode and cathode reactions. Cue b) Now show the overall cell reaction RA9 Coe f nKA c If [cu 10 M and [Ag -0.5M, calculate the cell voltage. Hint: Standard electrode potentials are absolute. They do not change if you change the stoichiometry of the reaction. Electrode Reaction E 46 V 2+ 2e Cu +0.34 Ag e Ag +0.80 Note: E Eo 2.303 RT log roducts nF reactants

Explanation / Answer

Cu ----------> Cu+2 + 2e-             E0 = -0.34v

2Ag+ + 2e- ----------> 2Ag          E0   = 0.80v

----------------------------------------------------------------------------------------

Cu(s) + 2Ag+ (aq) ---------> Cu+2 (aq) + 2Ag(s)    E0cell   = 0.46v

          n = 2

E cell     =   Ecell - 0.0592/n logQ

             = Ecell - 0.0592/n log[Cu+2]/[Ag+]2

                  =   0.46 - 0.0592/2 log0.001/(0.5)^2

             = 0.46-0.0296*-log0.00 4

            = 0.46-0.0296*-2.3979   = 0.531v >>>>>answer

          

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