I am confused one what principle to use and how to do also solve 61 when voltz a
ID: 1019106 • Letter: I
Question
I am confused one what principle to use and how to do also solve 61 when voltz are already given to me? 59. Use the table of standard reduction potentials (Table 18.1) to pick a reagent that is capable of each of the following oxidations (un- der standard conditions in acidic solution) a. oxidize Br" to Br but not oxidize Cl" to Cl [Ag+)-1,0 M b. oxidize Mn to Mn2 but not oxidize Ni to Ni 60. Use the table of standard reduction potentials (Table 18.1) to pick a reagent that is capable of each of the following reductions (un- reductions (un 66. Consider a 66. Consider a concentration cell similar to the one shown in Exer- der standard conditions in acidic solution) a. reduce Cu2+ to Cu but not reduce Cu2 to Cu b. reduce Br to Br but not reduce I2 to 1 cise 65, except that both electrodes are made of Ni and in the left-hand compartment [Ni2+] 1.0 M. Calculate the cell po- tential at 25°C when the concentration of Ni2+ in the compart- ment on the right has each of the following values. a. 1.0 M b. 2.0 M c. 0.10 M d. 4.0 x 10-5 M e. Calculate the potential when both solutions are 2.5 Min Ni2* For each case, also identify the cathode, anode, and the direc- tion in which electrons flow. 61) Hydrazine is somewhat toxic. Use the half-reactions shown be- low to explain why household bleach (a highly alkaline solutiorn of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. ClO-+ H2O + 2e- 2OH-+ Cl- go-0.90 V N2H, + 2H,0 + 2e-2NH, + 20H- W"=-0.10 V 62. The compound with the formula TIl, is a black solid. Given the 67 The overall reaction in the lead storage battery is following standard reduction potentials Pb(s) + PbO2(s) + 2H+ (aq) + 2HSO. (aq) 2PbSO(s) + 2H,00 h" + 2e-- 31- go-0.55V would you formulate this compound as thallium(H) iodide or thallium() triiodide? Calculate % at 25°C for this battery when [H2S 4.5 M, that is, [H+]s [HSO= 4.5 M. At 25°C,V' = 2.04 V for the lead storage battery
Explanation / Answer
61. The given half reactions are
ClO- + H2O + 2e- -----à 2 OH- + Cl- ; E0 = 0.90 V
N2H4 + 2 H2O + 2e- ----à 2 NH3 + 2 OH-; E0 = -0.10 V
The given values are standard electrode potentials. The second reaction is the reduction of hydrazine to ammonia. Remember that the more negative the value of the standard reduction potential, the greater is the tendency of the reaction to shift toward the product side, i.e, favour the reactants. In this case, we shall have more of hydrazine. This is similar to oxidation of ammonia to hydrazine. The oxidation reaction is the reverse of the given reduction reaction. Remember that when we reverse the reaction, the sign of E0 changes. The oxidation reaction is
2 NH3 + 2 OH- --------à N2H4 + 2 H2O + 2e-; E0ox = 0.10 V
Thus, when bleach comes in contact with ammonia, bleach oxidizes ammonia to hydrazine. The reaction takes place through the formation of intermediate chloramine, NH2Cl. Both chloramine and hydrazine are highly toxic. Exposure to hydrazine can cause irritation of eyes, nose, nausea, temporary blindness, dizziness, etc. Extreme effects of hydrazine poisoning can lead to seizures and coma. Hydrazine is also harmful for liver, kidneys and the central nervous system. Thus, bleach should never be allowed to come in contact with ammonia. The low standard reduction potential of hydrazine ensures easy oxidation of ammonia to hydrazine and hydrazine is severely harmful.
63. The standard reduction potential for the lead storage battery is E0 = 2.04 V.
The cell potential at 25C = 298 K is
Ecell = E0 – (RT/nF)ln[1/{[H+]2[HSO4-]2}]
Here, R = 8.315 J/mol.K, T = 298 K, n = 2 moles of electrons (total 2 moles of electrons are transferred here) and F = 96485 Coulombs. Thus,
Ecell = 2.04 – [(8.315)(298)/(2)(96485)]ln[1/(4.5)2(4.5)2] = 2.04 – (0.01284)ln(0.00244) = 2.04 – (0.01284)(-6.016) = 2.04 + 0.0772 = 2.117 2.12
The emf of the lead storage battery is 2.12 V (ans).
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