Question

In the electrochemical cell:

$Zn|ZnS{O}_4\left(0.01M\right)\parallel CuS{O}_4\left(1.0M\right)|Cu$, the emf of a Daniell cell at $298K$ is $E_1$. When the concentration of $ZnS{O}_4$ is $1.0M$ and that of $CuS{O}_4$ is $0.01M$, the emf changed to $E_2$. What is the relations between $E_1$ and $E_2$?

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

Answer 1

The correct option is C $E_1>E_2$

The concentration of reactants and products is interchanged. Hence, the value of the reactant quotient $Q$ will become reciprocal of the initial value.

$E_{cell}=E_{cell}^0-\frac{0.0592}{n}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
$E_1=E_{cell}^0-\frac{0.0592}{2}log\frac{0.01}{1.0}$

$E_1=E_{cell}^0+0.0592$ V ----- 1
$E_2=E_{cell}^0-\frac{0.0592}{2}log\frac{1.0}{0.01}$

$E_2=E_{cell}^0-0.0592$ V

Since $E_{cell}^o$ is same in both cases.
Hence, $E_1>E_2$