A) 1.10V
B) 0.42V
C) 0.84V
D) 1.34V
Correct Answer: A
Solution :
Anode is always the site of oxidation thus anode half cell is \[Z{{n}^{2+}}(aq)+2{{e}^{-}}\xrightarrow{{}}Zn(s);\]\[{{E}^{o}}=-0.76\,V\] Cathode hall cell is \[A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\xrightarrow{{}}\]\[2\,Ag(s)+2O{{H}^{-}}(aq);{{E}^{o}}=0.34\,V\] \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] \[=0.34-(-0.76)=+\,1.10\,V\]You need to login to perform this action.
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