Question

NUEL UILCO CHM 112 Lab Analysis 1 CHE [References Use the References to access important values if needed for this question. Enter electrons as c Use smallest possible integer coefficients. If a box is not needed, leave it blank. A voltaic cell is constructed from a standard Co2"Co half cell (Erod - -0.280V) and a standard APAl half cell (Ered -- 1.660V). The anode reaction is: + The cathode reaction is: + + The spontaneous cell reaction is: + The cell voltage is V. Submit Answer Try Another Version 4 item attempts remaining

Answer 1

• voltué cell is constructed from two half cell- 0824/20 (E® sud = -0.280 V) Al3+ / Al (Ested - - 1.660 V The anode reaction oxidation reaction Al The cathode reaction = Reduction reaction Al3+ +3e 2+ +2e - Co + 2+ Overall spontaneous cell reaction 2Al 2Al3+ +be 3 eo 2 + +66 300 cell 2A1+300 3+ 2Al + 300 reaction The given cell reaction 3 Ag + cool +Cr (s) 3 Ag (5) + eu 3+ (an) Standard reduction potential of two ب زر half cell is و موم * ہوم Ag + 1 Ag , Ered = 0.80 V Eored= -0.74 v cathode reduction reaction Agtte Ag Anode oxidation ca 3++3e cell = Eo cathode = E Agt/Ag 93tler 0.80 V-(-0.74 V - 6-80 +674) E' cell 1.54 v 0 E anode -EO

free energy change, AGO=-nf E' cell Gº=-3X 96500 X 1,54 --445830 J 460 = -445.83 We know that AGO = - RT en K or +445830 = 78.3148 2988 link or en k = 415830 8,314x298 org enk 179,946 1. AIX1078 k for this reaction would be greater than one. The given reaction is 2Ag + C1-17 M) ted (s) + 2Ag (s) + 6d2+ (5.91810-3M) AG-245 KT The reduction potential of the two half cells Agt lag Eorud 0-80 v ed²tled, Ered = -0.40 v Eo cell to cathode -€° anode 0.80 - (-0.40) 1.2 According to Nernst equation E = €° cell 2.303 RT log [ed 2+ [Ag +] 2 - 1.2- 2:303 X 8.314 X 298 (5.91 X10-3) 2X 96500 (117) - 1.2 - 0.0295 log (4.317 X10-3) Er 12+ 0.0697 E 1.269 v E' cell F log

For spontaneous reaction AG Value should be negative and E value should be positive. As here AG -245 KZ which is negative value, the reaction is spontaneous in forward direction