The following half reactions appear in the table of the information page Br_2(I)
ID: 521508 • Letter: T
Question
Explanation / Answer
Ag (s)<--->Ag+ (aq) +e- (1), Eo= -0.80 V
Br2+2e<-->2Br- (2), Eo=1.066
Multiplying eq.1 with 2 and addition with Eq.2 gives
2Ag(s) +Br2(l) <---->2Ag+ (aq)+2Br-, Eo= 1.066-0.8= 0.266V
If delaG is -ve, the reaction is spontaneous.
deltaG=-nFE, n= no of electrons exchanged= 2, F= 96500 and E= 0.266V
deltaG=-2*96500*0.266=-51338 joules
2.
F2+2e- -------->2F- , Eo= 2.87V (1)
Co(s) ---------->Co+2+2e-, Eo= 0.288V (2)
addition of two reactions give F2+Co -------->2F- +CO+2, Eo= 2.87+0.288= 3.158 V
deltaG=-2*96500*3.158 =-609494 joules, spontaneous.
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