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Univ × ww.saplinglearning.com/ibiscms/mod/ibis/view.php?id- 3850320 rning tic Activities and Due Dates HW 21 lm Beach Atlantic University- CHM 1023-Fall17-GRANT Gradeboc 12/2/2017 11:55 PM 07.1/1011/28/2017 10:53 PM to-, Print Calalator st Periodic Table core Question 8 of 20 Map Sapling Learning A current of 4.58 A is passed through a Fe(NOsh solution. How long (in hours) would this current have to be applied to plate out 5.60 g of iron? Number O Previous Gwe up & View Scation eCheck Answer Next Exe

Explanation / Answer

Fe^2+ (aq) + 2e^- ------------> Fe(s)
         Z = 2
         C = 4.58A
         F = 96500c
         W = 5.60g
         M   = 56g/mole
     W   = MCt/ZF
     t   = WZF/MC
         = 5.6*2*96500/56*4.58   = 4213.97sec
    [1h   = 60*60sec = 3600sec]
         = 4213.97/3600   = 1.17h >>>>>answer
Q2
       Mg(s) ---------> Mg^2+ (aq) + 2e^-      E0 =2.38v
       Sn^2+ (aq) + 2e^- ------> Sn(s)         E0 =-0.14v
      --------------------------------------------------------------
        Mg(s) + Sn^2+ (aq) -----> Mg^2+ (aq) + Sn(s) E0cell = 2.24v
           n = 2
      Ecell   = E0cell - 0.0592/n log[Mg^2+]/[Sn^2+]
              = 2.24- 0.0592/2 log0.806/0.011
              = 2.24-0.0296*1.8649
               = 2.185v >>>>answer
     Zn(s) ----------> Zn^2+ (aq) + 2e^-     E= 0.76v
     Zn^2+ (aq) + 2e^- ----> Zn(s)          E0 = -0.76v
     --------------------------------------------------
     Zn(s) + Zn^2+ (aq) --------> Zn^2+ (aq) + Zn(s)   E0cell =0
        n =2
     Ecell = E0cell - 0.0592/n log[Zn^2+]/[Zn^2+]
     0.01   = 0     - 0.0592/2 log0.1/[Zn^2+]
     0.01   = -0.0296log0.1/[Zn^2+]
     log0.1/[Zn^2+] = -0.01/0.0296
     log0.1/[Zn^2+] = -0.3378
        0.1/[Zn^2+]   = 10^-0.3378
        0.1/[Zn^2+]   =0.4594
            [Zn^2+]   = 0.1/0.4594   = 0.2176M
   

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