Question

Consider the cell : $A{g}_{\left(s\right)}|AgCl\left(\text{saturated solution}\right)|AgN{O}_{3\left(aq\right)}1.0M|A{g}_{\left(s\right)}$

EMF of the above cell is given by:

Given: $(K_{sp}$ of $AgCl$ $=1.0\times {10}^{-10}{M}^2)$

• A. $E_{cell}=0.059log\left[K_{sp}\right(AgCl){]}^{1/2}$
• B. $E_{cell}=0.059log\frac{1}{\left[K_{sp}\right(AgCl){]}^{1/2}}$
• C. $E_{cell}=0.059\times 5V$
• D. $E_{cell}=296mV$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $E_{cell}=0.059log\frac{1}{\left[K_{sp}\right(AgCl){]}^{1/2}}$
C $E_{cell}=0.059\times 5V$
D $E_{cell}=296mV$

For anode,

$K_{sp}=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]={\left[{Ag}^{+}\right]}^2$

$\left[{Ag}^{+}\right]=\sqrt{K_{sp}}$

For cathode, $\left[{Ag}^{+}\right]=1.0M.$

Hence the equilibrium constant will be $K=\sqrt{K_{sp}}$

The standard emf of the cell is $0.0V$.

The emf of the cell will be

$E_{cell}=E_{cell}^0-\frac{0.0592}{n}logK=0.0-\frac{0.0592}{1}log\sqrt{K_{sp}}=0.0592log\frac{1}{\sqrt{K_{sp}}}$

Substitute values in the above expression

$E_{cell}=0.0592log\frac{1}{\sqrt{1\times {10}^{-10}}}=0.0592\times 5V=0.296V=296mV.$