Question

For the galvanic cell :

Ag 1 AgCl(s), KCl(0.2 M)II KBr (0.001 M), AgBr(s) I Ag, the EMF (in V) generated for a spontaneous process after taking into account the cell reaction at ${25}^{\circ }C$ will be $37\times {10}^{-x}$, then what is the value of x?

$\left[K_{sp}\right(AgCl)=2.8\times {10}^{-10},K_{sp}AgBr)=3.3\times {10}^{-13}]$

Answer 1

For the galvanic cell,

Ag I AgCl(s) . KCl (0.2M) II KBr (0.001 M) AgBr(s) I Ag

$E_{cell}=\frac{0.591}{n}log\frac{[A{g}^{\oplus }{]}_{AgBr}}{[A{g}^{\oplus }{]}_{AgCl}}$

$K_{sp}ofAgCl=\left[A{g}^{\oplus }\right]\left[C{l}^{\ominus }\right]$

For 0.2 M KCl,
$\left[C{l}^{\ominus }\right]=0.2M$

$2.8\times {10}^{-10}=\left[A{g}^{\oplus }\right]\times 0.2$

Or $[A{g}^{\oplus }{]}_{AgCl}=\frac{2.8\times {10}^{-10}}{0.2}=1.4\times {10}^{-10}M$

$K_{sp}ofAgBr=\left[A{g}^{\oplus }\right]\left[B{r}^{\ominus }\right]$

From 0.001 M KBr

$\left[B{r}^{\ominus }\right]$ = 0.001 M

$\therefore 3.3\times {10}^{-13}=\left[A{g}^{\oplus }\right]\times 0.001M$

Or $[A{g}^{\oplus }{]}_{AgBr}=\frac{3.3\times {10}^{-13}}{0.001}=3.3\times {10}^{-10}M$

From equiation (i),

$E_{cell}=\frac{0.0591}{1}log\frac{3.3\times {10}^{-10}}{14\times {10}^{-10}}$

$=0.0591[log3.3\times {10}^{-10}-log14\times {10}^{-10}]$

$=0.0591[0.5187101.1461+10]$

$=0.0591(-0.6274)=-0.037V$

Thus, to get $E_{cell}$ positive, the polarity of the cell should be reversed, i.e. cell is Ag I AgBr(s) (0.001 M KBr) II AgCk (s) (0.2 M KCl) I Ag

So, $E_{cell}$ = $+0.037\approx 37\times {10}^{-3}V$