Given cell : $$Ag|A{g^ + }(sat\,A{g_2}Cr{O_4}\,sol)||Ag(0.1\,M)|Ag$$ and $${E_{cell}} = 0.164\,V$$ at 298 K.

At anode : $$Ag(s) \to A{g^ + }(sat\,A{g_2}Cr{O_4}) + e$$

At cathode : $$A{g^ + }(aq) + e \to Ag(s)$$

Cell reaction : $$A{g^ + }(aq) \to A{g^ + }(sat\,A{g_2}Cr{O_4})$$

Let molar concentration of Ag⁺ ions in sat Ag₂CrO₄ be C₁M

$$\therefore$$ $${E_{cell}} = E_{cell}^0 - {{0.059} \over 1}\log {{{C_1}} \over {[A{g^ + }]}}$$

$$E_{cell}^0 = 0$$ as both the electrodes are the same

$$\therefore$$ $$0.164 = {{0.059} \over 1}\log {{0.1} \over {{C_1}}}$$ or, $$\log {{0.1} \over {{C_1}}} = {{0.164} \over {0.059}} = 2.774$$

or, $${C_1} = {[A{g^ + }]_{A{g_2}Cr{O_4}}} = 1.66 \times {10^{ - 4}}(M)$$

In, $$A{g_2}Cr{O_4}(s)$$ $$\rightleftharpoons$$ $$2A{g^ + }(aq) + CrO_4^{2 - }(aq)$$

$$[CrO_4^{2 - }] = {{[A{g^ + }]} \over 2} = {{1.66 \times {{10}^{ - 4}}} \over 2}(M) = 0.83 \times {10^{ - 4}}(M)$$

$$\therefore$$ K_sp of $$A{g_2}Cr{O_4} = {[A{g^ + }]^2}[CrO_4^{2 - }]$$

$$ = (1.66 \times {10^{ - 4}}) \times 0.83 \times {10^{ - 4}} = 2.287 \times {10^{ - 12}}$$