A) \[~+\text{ }1.10\,V\]
B) \[-1.10\,V\]
C) \[-0.42\,V\]
D) \[+\,0.42\,V\]
Correct Answer: A
Solution :
\[Zn\xrightarrow{{}}Z{{n}^{2+}}+2{{e}^{-}}\] (oxidation half-reaction) \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\](reduction half-reaction) Emf of the cell = standard oxidation potential of zinc electrode + standard reduction potential of Cu electrode \[=0.76\text{ }V+0.34\,V\] \[=1.10\,V\]
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