Question

If the solution of copper sulphate in which a copper rod is immersed, is diluted 100 times, what is the change in electrode potential (Reduction)?

• A. $ -\text{ }29.5\text{ }mV $
• B. $ 29.5\text{ }mV $
• C. $ -59.0\,mV $
• D. $ 59.0\text{ }mV $

Answer 1

Answer: (D)

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.