For the galvanic cell,

$$Ag|AgCl(s)$$, $$KCl(0.2M)||KBr(0.001\,M)$$, $$AgBr(s)|Ag$$

$${E_{cell}} = {{0.591} \over n}\log {{[A{g^ + }]AgBr} \over {[A{g^ + }]AgCl}}$$ ...... (1)

Given, $${K_{sp}}(AgCl) = 2.8 \times {10^{ - 10}}$$. Now, [Cl^$$-$$] = 0.2 (M)

$$\therefore$$ $$2.8 \times {10^{ - 10}} = {[A{g^ + }]_{AgCl}} \times 0.2$$

or, $${[A{g^ + }]_{AgCl}} = {{2.8 \times {{10}^{ - 10}}} \over {0.2}} = 14 \times {10^{ - 10}}(M)$$

Given, $${K_{sp}}(AgBr) = 3.3 \times {10^{ - 13}}$$. Here, [Br^$$-$$] = 0.001 M

$$\therefore$$ $$3.3 \times {10^{ - 13}} = {[A{g^ + }]_{AgBr}} \times 0.001$$

or, $${[A{g^ + }]_{AgBr}} = {{3.3 \times {{10}^{ - 13}}} \over {0.001}} = 3.3 \times {10^{ - 10}}$$ (M)

Substituting in equation (1), $${E_{cell}} = {{0.059} \over 1}\log {{3.3 \times {{10}^{ - 10}}} \over {14 \times {{10}^{ - 10}}}}$$

= 0.059 $$\times$$ ($$-$$ 0.6274) = $$-$$ 0.037 V

The negative value of E_cell indicates that the cell reaction $$[AgBr(s) + C{l^ - }(aq) \to AgCl(s) + B{r^ - }(aq)]$$ is not spontaneous but the reverse reaction is spontaneous as in this case E_cell = +0.037 V. To carry out the reverse reaction, the polarity of the given cell should be reversed i.e., the anode should be made cathode, and vice-versa.