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. $-29.5mV$
• B. $29.5mV$
• C. $-59.0mV$
• D. $59.0mV$

Answer 1

Answer: (D)

For reduction process, $\underset{xM}{C{u}^{2+}}+2e^{-}\to Cu$ $E_{C{u}^{2+}/Cu}=E_{C{u}^{2+}+Cu}^o-\frac{0.0591}{2}\log \frac{1}{x}$ ?(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 $E_{C{u}^{2+}/Cu}^{}-E_{C{u}^{2+}/Cu}$ $=-\frac{0.0591}{2}\log \frac{{10}^2}{x}+\frac{0.0591}{2}\log \frac{1}{x}$ $=\frac{0.0591}{2}\left[\log \frac{{10}^2}{x}-\log \frac{1}{x}\right]$ $=-\frac{0.0591}{2}[\log {10}^2-\log x+\log x]$ $=-\frac{0.0591}{2}\log {10}^2=-\frac{0.0591}{2}\times 2$ $=-0.0591V=-59.1mV$ Thus, electrode potential decreases by 59.0 mV.