Question

If $Z{n}^{2+}/Zn$ electrode is diluted 100 times, then the charge in reduction potential is

• A. increase of 59 mV
• B. decrease of 59 mV
• C. increase of 25.5 mV
• D. decrease of 2.95 mV

Answer 1

Answer: (B)

$E_{Z{n}^{+}/Zn}=E_{Z{n}^{2+}/Zn}^0-\frac{0.059}{2}log\frac{1}{|Z{n}^{2+}|}$
$E_{Z{n}^{+}/Zn}=E_{Z{n}^{2+}/Zn}^0+0.0295logC$
When solution is diluted $100$ times
$E_{Z{n}^{2+}/Zn}^{\prime }=E{°}_{Z{n}^{2+}/Zn}+0.0295log\frac{C}{100}$
$=E{°}_{Z{n}^{2+}/Zn}+0.0295logC-0.0295\times 2$
$=E_{Z{n}^{2+}/Zn}-0.059$
$E_{Z{n}^{2+}/Zn}^{\prime }-E_{Z{n}^{2+}/Zn}=-0.059V$
$=-59mV$
Thus $R.P$. will decrease by $59mV$.