Question

In the electrochemical cell:
$Zn|S{O}_4\left(0.01M\right)||CuS{O}_4\left(1.0M\right)|Cu$, the emf of this Daniel cell is $E_1$. When the concentration of $ZnS{O}_4$ is changed to $1.0M$ and that of $CuS{O}_4$ is changed to $0.01M$, the emf changes to $E_2$. From the followings, which one is the relationship between $E_1$ and $E_2$?
$(\text{Given},\frac{RT}{F},0.059)$

• A. $E_1<E_2$
• B. $E_1>E_2$
• C. $E_2=0\ne E_1$
• D. $E_1=E_2$

Answer 1

The correct option is B $E_1>E_2$

Cell reaction:
$Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$
$E_1=E^{\circ }+\frac{0.0591}{2}lo{g}_{10}\frac{\left[C{u}^{+2}\right]}{\left[Z{n}^{+2}\right]}$
$E_1=E^{\circ }+0.03lo{g}_{10}\left[\frac{1}{0.01}\right]$
$E_1=E^{\circ }+0.06$

$E_2=E^{\circ }+\frac{0.0591}{2}lo{g}_{10}\frac{\left[Z{n}^{+2}\right]}{\left[C{u}^{+2}\right]}$
$E_2=E^{\circ }+0.03lo{g}_{10}\left[\frac{0.01}{1}\right]$
$E_2=E^{\circ }-0.06$
From above, $E_1>E_2$