NEET UG 2022 — Chemistry Physical Chemistry

Given reaction: $\mathrm{Ni}(s)+2{\mathrm{Ag}}^{+}(0.001M)\rightarrow {\mathrm{Ni}}^{2+}(0.001M)+2\mathrm{Ag}(s)$Ni(s)+2Ag+(0.001M)→Ni2+(0.001M)+2Ag(s)

Applying Nernst equation we have:

$${E}_{(\mathrm{cell})}={E}_{(\mathrm{cell})}^{0}-\frac{2.303\mathrm{RT}}{\mathrm{nF}}\mathrm{log}\frac{[{\mathrm{Ni}}^{2+}][\mathrm{Ag}]}{{[{\mathrm{Ag}}^{+}]}^{2}[\mathrm{Ni}]}$$E(cell)​=E(cell)0​−nF2.303RT​log[Ag+]2[Ni][Ni2+][Ag]​

Active mass of solid is taken to be unity so $[\mathrm{Ni}(s)]=[\mathrm{Ag}(s)]=1$[Ni(s)]=[Ag(s)]=1

$${E}_{(cell)}={E}_{(cell)}^{0}-\frac{0.059}{n}\mathrm{log}\frac{[N{i}^{2+}]}{{[A{g}^{+}]}^{2}}$$E(cell)​=E(cell)0​−n0.059​log[Ag+]2[Ni2+]​

$$=1.05-\frac{0.0591}{2}\mathrm{log}\frac{(0.001)}{{(0.001)}^{2}}$$=1.05−20.0591​log(0.001)2(0.001)​

$$=1.05-0.0295\mathrm{log}(1\times {10}^{3})$$=1.05−0.0295log(1×103)

$$=1.05-0.0295\times 3$$=1.05−0.0295×3

$$=0.9615V$$=0.9615V

Therefore, the emf of the cell is $0.9615V$0.9615V.

Note: Question is modified little bit for academic accuracy. $E{^{\circ}}_{\mathrm{cell}}=1.05V\mathrm{in}\mathrm{place}\mathrm{of}E{^{\circ}}_{\mathrm{cell}}=10.5V$E∘cell​=1.05VinplaceofE∘cell​=10.5V as given in NEET 2022 paper.