A) Copper electrode is work like cathode, then \[E_{cell}^{o}\] is \[+0.45V\]
B) Silver electrode is work like anode then \[E_{cell}^{o}\] is \[-0.34V\]
C) Copper electrode is work like anode then \[E_{cell}^{o}\] is \[+0.46V\]
D) Silver electrode is work like cathode then \[E_{cell}^{o}\] is \[-0.34V\]
E) Silver electrode is work like anode then \[E_{cell}^{o}\] will be \[+1.14V\]
Correct Answer: C
Solution :
Cell reaction is \[C{{u}_{(s)}}+2A{{g}^{+}}\to C{{u}^{2+}}+2Ag\] Two half cell reaction is \[Cu\to C{{u}^{2+}}+2{{e}^{-}}\] Oxidation (anode) \[A{{g}^{+}}+{{e}^{-}}\to Ag\] Reduction (cathode) \[{{E}_{Cell}}={{E}_{ox}}-{{E}_{\operatorname{Re}d}}=0.80-0.34=+0.46V\]
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