Question

In a Daniel cell constructed in the laboratory.

The voltage observed was 0.9V instead of 1.10V of the standard cell. A possible explanation is:

• A. molar ratio of $Z{n}^{2+}:C{u}^{2+}$is 2 : 1
• B. the Zn electrode has thrice the surface of Cu electrode
• C. $\lfloor Z{n}^{2+}\rfloor <\lfloor C{u}^{2+}\rfloor$
• D. $\lfloor Z{n}^{2+}\rfloor >\lfloor C{u}^{2+}\rfloor$

Answer 1

The correct option is D $\lfloor Z{n}^{2+}\rfloor >\lfloor C{u}^{2+}\rfloor$

$0.9=1.0-\frac{0.059}{2}ln\left[\frac{Z{n}^{2+}}{C{n}^{2+}}\right]$

$\therefore ln\left(\frac{Z{n}^{2+}}{C{u}^{2+}}\right)>0$

$\therefore \left[Z{n}^{2+}\right]>\left[C{u}^{2+}\right]$

Hence, option D is correct.