A) \[+1.10\,\,V\]
B) \[-1.10\,\,V\]
C) \[-0.76\]
D) \[-0.42\]
Correct Answer: A
Solution :
\[E_{cell}^{\text{o}}=E_{cathode}^{\text{o}}-E_{anode}^{\text{o}}\] Oxidation half reaction\[=Zn\xrightarrow{{}}Z{{n}^{2+}}2{{e}^{-}}\] \[\frac{\operatorname{Re}duction\,\,half\,\,reaction=C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu}{Cell\,\,reaction=Zn+C{{u}^{2+}}\xrightarrow{{}}2Z{{n}^{2+}}+Cu}\] \[E_{right}^{\text{o}}-E_{left}^{\text{o}}=E_{cell}^{\text{o}}\] \[=0.34-(-0.76)\] \[=0.34+0.76\] \[=+1.10\,\,V\]
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