Question

What will be the effect on emf of given cell when the concentration of $A{g}^{+}$ at cathode is decreased to $0.1M$?
$Ag\left(s\right)|A{g}^{+}(aq,0.001M)||A{g}^{+}(aq,1M)|Ag\left(s\right)$

• A. Emf of the cell gets reduced from $0.177$ to $0.118V$
• B. Emf of the cell increased from $0.117$ to $3.31V$
• C. Emf of the cell gets reduced from $0.214$ to $0V$
• D. Emf of the cell increased from $0.118$ to $0.214V$

Answer 1

The correct option is A Emf of the cell gets reduced from $0.177$ to $0.118V$

From cell representation,
$\left[A{g}^{+}\right]=0.001M\Rightarrow$ Anode
$\left[A{g}^{+}\right]=1M\Rightarrow$ Cathode

Cell reaction will be,
$A{g}^{+}(aq,1M)\to A{g}^{+}(aq,0.001M)$

By nernst equation,

$E_{cell}=E_{cell}^0-\frac{0.0591}{1}log\frac{\left[0.001\right]}{\left[1\right]}$

$E_{cell}=E_{cell}^0+\frac{0.0591}{1}log\frac{\left[1\right]}{\left[0.001\right]}$

For concentration cells, cathode and anode consist of same metal and its solution​.
Thus,
$E^0=0$

$E_{cell}=\frac{0.0591}{1}log\frac{\left[1\right]}{\left[0.001\right]}$

$E_{cell}=\frac{0.0591}{1}log{10}^3$

$E_{cell}=\frac{0.0591}{1}\times 3$

$E_{cell}\approx 0.177V$

Now, concentration of $A{g}^{+}$ at cathode is reduced to $0.1M$

Cell reaction will be,
$A{g}^{+}(aq,0.1M)\to A{g}^{+}(aq,0.001M)$

$E_{cell}=E_{cell}^0-\frac{0.0591}{1}log\frac{\left[0.001\right]}{\left[0.1\right]}$

$E_{cell}=E_{cell}^0+\frac{0.0591}{1}log\frac{\left[0.1\right]}{\left[0.001\right]}$

$E^0=0$

$E_{cell}=\frac{0.0591}{1}log\frac{\left[0.1\right]}{\left[0.001\right]}$

$E_{cell}=\frac{0.0591}{1}log{10}^2$

$E_{cell}=\frac{0.0591}{1}\times 2$

$E_{cell}\approx 0.118V$