A) \[+1.10\text{ }V\]
B) \[-1.10\text{ }V\]
C) \[+\text{ }0.42V\]
D) \[-0.42V\]
Correct Answer: A
Solution :
Key Idea: (i) Find cathode and anode from given cell reaction. (ii) Calculate emf by formula \[{{E}_{cell}}={{E}_{c}}-{{E}_{a}}\] Given that \[Zn/Z{{n}^{2+}}||C{{u}^{2+}}/Cu\] \[\therefore \]Zn is anode and Cu is cathode Given, \[Z{{n}^{2+}}/Zn=-0.76\,V\] \[C{{u}^{2+}}/Cu=+0.34\,V\] \[{{E}_{cell}}={{E}_{c}}-{{E}_{a}}\] \[=0.34-(-0.76)\] \[=0.34+0.76\] \[=1.10V\]You need to login to perform this action.
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