Please answer the following question and show all work to receive credit:
Homework Answers
anode : oxidation takes place
Zn (s) -------------------> Zn+2 (aq) + 2e , E0Zn+2/Zn = -0.76
cathode : reduction takes place
Br2 (aq)+ 2e- --------------------> 2 Br- (aq) , E0Br-/Br2 = +1.07
net reaction :
Zn (s) + Br2 (aq) -----------------> Zn+2 (aq) + 2 Br- (aq)
E0cell= E0cathode- E0anode
E0cell= E0Br-/Br2 - E0Zn+2/Zn
= 1.07 - (-0.76)
= 1.83 V
E0cell = 1.83 V
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Please answer the following question and show all work to
receive credit:
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