A) \[-\text{ }29.5\text{ }mV\]
B) \[29.5\text{ }mV\]
C) \[-59.0\,mV\]
D) \[59.0\text{ }mV\]
Correct Answer: D
Solution :
For reduction process, \[\underset{xM}{\mathop{C{{u}^{2+}}}}\,+2{{e}^{-}}\xrightarrow{{}}Cu\] \[{{E}_{C{{u}^{2+}}/Cu}}=E_{C{{u}^{2+}}+Cu}^{o}-\frac{0.0591}{2}\log \frac{1}{x}\] ?(i) When the solution is diluted 100 times, \[x=\frac{x}{100}\] \[E_{C{{u}^{+}}/Cu}^{}=E_{C{{u}^{2+}}/Cu}^{o}-\frac{0.0591}{2}\log \frac{{{10}^{2}}}{x}\] ?(ii) On subtracting Eq (ii) -Eq (i), we get \[E_{C{{u}^{2+}}/Cu}^{}-{{E}_{C{{u}^{2+}}/Cu}}\] \[=-\frac{0.0591}{2}\log \frac{{{10}^{2}}}{x}+\frac{0.0591}{2}\log \frac{1}{x}\] \[=\frac{0.0591}{2}\left[ \log \frac{{{10}^{2}}}{x}-\log \frac{1}{x} \right]\] \[=-\frac{0.0591}{2}[\log {{10}^{2}}-\log x+\log x]\] \[=-\frac{0.0591}{2}\log {{10}^{2}}=-\frac{0.0591}{2}\times 2\] \[=-0.0591V=-59.1\,mV\] Thus, electrode potential decreases by 59.0 mV.
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