Please explain your answer! (just 3 & 4) 3. The cell potential of an electrochem
ID: 866966 • Letter: P
Question
Please explain your answer! (just 3 & 4)
3. The cell potential of an electrochemical cell with the cell reaction Al(s Crs (aq) Cr(s) AP (aq) is 1.63 V. What is the maximum electrical work obtainable from this cell when 0.50 g of Al is consumed? A) -1.7x 104 J B) -2.4 x 103 J C) -8.7 x 103 J D) -2.9 x 103 J E) -6.4 x 10% J 4. Which reaction would be most likely to occur at the anode of a voltaic cell? A) PbSO4 (s) 2e Pb(s) SO42 (aq) C) PbSO4 (s) Pb2 (aq) SO42 (aq) E) 2H20 2e H2(g) 20H (aq) 5. What is Eocell for the cell reaction 2Al(s)+ 3sn' (aa) 3sn? (aa) +2AP (aq)? 1.66V sn' (aa)+2e E)) 1.81V D) 1.93V C) 1.51 V B) -1.51 V A) 0.45 V What is the balanced spontaneous reaction and standard cell potential of an electrochemic constructed from half cells with the following half reactions? Eo -0.763 VExplanation / Answer
3. The given cell reaction is
Al(s) + Cr3+ (aq) ---------------> Cr (s) + Al3+ (aq) , E = 1.63 V
The half cell reactions can be written as
Al(s) ------------> Al3+ (aq) + 3e-
Cr3+ (aq) + 3e- -------> Cr (s)
Hence number of moles of electrons transferred during the reaction by 1 mol of Al is n = 3 mol.
Now maximum work done by an electrochemical cell is given as
Wmax = - nFE = - 3 mol x 96500Cmol-1 x 1.63 V = 471885 CxV = - 471885 J [since CxV = Jule]
The above calculated work done is for 1 mole of Al (= 27 g Al)
Therefore maximum work done by 0.5 g of Al = ( - 471885 J/ 1 mol Al) x( 0.5 g Al)x (1 mol Al / 27 g Al)
- 8738.6 J = - 8.7 x 103 J (answer) . Hence (C) is the correct answer.
Maximum work done by a cell is given by
Wmax = nFEcell
4. Oxidation takes place at the anode of a voltaic cell releasing electron. Among the given options in
(A) Pb2+ in the reactant is reduced to Pb in the product. Hence this is not the correct answer.
(B) Here 2H2O(l) is oxidised to O2(g) releasing 4 electrons. Hence this is the correct anodic reaction.
(C) Here neither oxidation nor reduction occurs. Hence this is also not the correct answer.
(D) Here 2H2O(l) is both oxidised and reduced to O2 and H2 respectively. Hence this is also not the correct answer.
(E) Here 2H2O(l) is reduced to H2 by receiving electron. Hence this is also not the correct answer
Hence (B) is the correct anodic reaction in voltaic cell.