Question

For the cell reaction:

$Zn\left(s\right)+C{u}^{2+}\left(0.1M\right)\to Z{n}^{2+}\left(0.01M\right)+Cu\left(s\right)$

If $E^0$ is the standard emf of the cell at 298 K, then:

• A. $E=E^0$
• B. $E>E^0$
• C. $E<E^0$
• D. $E=0$

Answer 1

The correct option is B $E>E^0$

$E=E^0-\frac{0.059}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

$=E^0-\frac{0.059}{2}\log \frac{0.01}{0.1}$

$=E^0-\frac{0.059}{2}\log \frac{1}{10}$

$E=E^0+0.0295$

$\therefore E>E^0$