A) \[2\]
B) \[100\]
C) \[{{10}^{-2}}\]
D) \[1\]
Correct Answer: C
Solution :
\[Zn+C{{u}^{2+}}\xrightarrow{{}}Z{{n}^{2+}}+Cu\] \[{{E}_{cell}}-{{E}^{o}}_{cell}=-\frac{0.0591}{2}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] i.e. \[0.0591=-\frac{0.0591}{2}\log \,\frac{{{C}_{1}}}{{{C}_{2}}}\] \[\log \frac{{{C}_{1}}}{{{C}_{2}}}=-2\] \[~\frac{{{C}_{1}}}{{{C}_{2}}}=Antilo{{g}^{2}}={{10}^{-2}}\]You need to login to perform this action.
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