Question

The following problems are based upon the following voltaic cell (concentration of aqueous solutions are 1 M): (Show all work for full credit.) Cr(s) c (aq) IlCut (aq) Cu (s) (3) I. From the voltaic cell written above, which half cell will be the anode, and which will be the cathode? Why? (3) 2. Write a balanced equation describing the redox reaction in this cell. (Molecular or Net lonic equation is correct) (2) 3. How many electrons are transferred in the balanced electrochemical reaction? (1) 4. Do you expect this reaction to be spontaneous? What should the sign of AG be? (6) 5. For Part II, your choices of voltaic cells include tin, copper, aluminum, and zinc. For the following voltaic cells, determine which would be the anode and which would be the cathode: Cu + Sn: Sn + AI Cu + AI: Sn + Zn Cu + Zn: Al + Zn (6) 6. For each of the six pairs of half cells above, record the number of electrons that would be transferred in a balanced reaction between them.

Answer 1

Cell notation : cro. 1 Costag, Il Cutag, 1CwG) 3 ) In Upac representation of voltaic cell, anode is shown left and cathode is at right. Hene anode is cool Costag (Co electrode) Cathode is custa Cu (Cu electrode) 2 Anode reaction is oxidation ie Loss of ē Coco ( 3 + 0 0 -0.741] Cathode reaction is reduction in Gain of ē cu trag) + 2 Curss (E = 0.34 v) No. of ē on both half oxns should be equal. To get the balanced equation, multiply equo by 2 and eqn ③ by 3, and add the resultant equation 26C75) 2003+ +62 3 cutlaq, +6 ł 3 cus) acr6 + 3 cuat +60 + 2 C 3 + 3 Cust be cancel b e from both sides. 2 Cris) + 3 custa 2023 tag + 3 Cus) 3 Number of elections transferred = 6 4 A cell xxn is spontaneous if Ecall is positive E cell = o Cathode, - E anode = 0.34v - -0.74 V = 0.34v + 0740 = 1.08V E = + 1.08v, hence reaction is spontaneous DDG m will be negative, AG = - RFE 9) +

AG = -6x96487 (I) (1:08 V) = 625,235.76 Jx1KJ) 1000 = -625.2 KJ -1-66 V 6 5) It is already mentioned that Evalue should be positive for a cell to function spontaneously. Ellen = Es Cathude] Eanodes - tue if E cathode higher than value than Eanodes a) cutsn; E Gath = 0.340; Esnt/sn = -0.14V. (higher éo) Clower E) Dam Sn is anode and cu is cathode. b) Cut Al; EOAPA = -1.660 Anode : Al » Cathode Cu. Cutzn ; to ren2t/n) = -0.76V Anode zn; Cathode cu d) Sn + Al., Ent/) = -0.14V ; E(AHA) - Lowers (higher) Anode : Al; Cathode :Sn e Sn + Zn ; Anode : Zn; Cathode: Sn. f Al + Zn Anode: Al; Cathode: Zn. 6) a) Balanced equ: To Sucs + cutegi sut + Cues, Number of ē = 2 b) An Al + curt Aßt cu do 201 + 350+ 2AP + + 3 Sn 2A1 + 3 cult 2013 + + 34 No. of ē = 6. No. of ē = 6 ) zu & Salt zu + Sn. 2 zn + cut + Cu No. of è = 2 . No. of ē = 12 2A1+ 3znt QAP+ 3 zn cell a b c d e f e = 6. No. of No. of ē 2 6 2 6 2 6 1 27