Question

(14-17) (a) Write the line notation for the following cell. reaction. (c) From you sometimes led (d) Extra cred manganese reactions in 2M HBr(aq, 0.10 M) Mn Use the me manganese net cell rea The m: (4). For ea ber of mol tion is AG in Append electroni Br (0) Al(NO )3(aq, 0.010 M) (b) Calculate the potential of each half-cell and the cell voltage, E. In which direction will electrons flow through the circuit? Write the spontaneous net cell reaction. (c) The left half-cell was loaded with 14.3 mL of Br (I) (density = 3.12 g/mL). The aluminum electrode contains 12.0 g of Al. Which element, Br, or Al, is the limiting reagent? (That is, which reagent will be used up first?) (d) If the cell is somehow operated under conditions in which it produces a constant voltage of 1.50 V, how much electrical work will have been done when 0.231 mL of Br(t) has been consumed? (e) If the potentiometer is replaced by a 1.20-k2 resistor and if the heat dissipated by the resistor is 1.00 X 10 J/s, at what rate (grams per second) is Al(s) dissolving? (In this question, the voltage is not 1.50 V.) 14-21. Su 0.10 Min (a) Usi 2014 and (b) In wl (C) Now 2Ag + you will n 14-22. The tion poten Pt(s) The te spheric p water is 727.2 Tor ficients, i 14-18. A nickel-metal hydride rechargeable battery formerly used in laptop computers is based on the following chemistry: Cathode: NiOOH(s) + H2O + e N i(OH)2(s) + OH discharge 334

Answer 1

(a) Pt(s) Bra(1) I HBr(aq, 0.10 M) || AI(NO3)2(aq, 0.010 M) | Al(s)

(b) RHC: A1+ 3e- Al(s) EE -1.677 V LHC: Bry() + le- = 2Br Eº: 4.078 V RHCE = -1.677- oostblog oé : 1.716V LHC: E.: +1.078 -0.0597617 10.10)' = 1.137V Er-E. = 2.854V RHCO, LHC (A1) a (pt) Redaction here { Brill) + ATC) = 3Br + AR 3+

m= 44.69 (c) 14.3 ml Brz e = 3.129 mi' ..0.279 ml Bra(e) 12.0g of Al@ anode 12.09 - 0.495 me 26.98 - gner 3½ ml Bra per 12 yonig noe Ime Al so 415 mil Br2 runs out/use up first!

(d)

E = 1.50V, wak ū 0.231 mL Br2 cansund 0.231 mL (3.12 g ml-1) = 159.8082 g molas 4.51 x 10-3 mil (Bra) 2 mole- / Imal Brz - q = (9.02210-3 mole-) (9.649x109 (melns) = 870C -W- Eię = (1.5V) (870) = 1.31x1031 = 1.31 kJ

According to HOMEWORKLIB POLICY only 1 question with 4 sub part per post is allowed. please post the rest question seperately.

Hope you understand!!

PLEASE LEAVE THE POSITIVE FEEDBACK IF YOU LIKE THE SOLUTION