QNEET-2022-Ch-395: NEET 2022 Chemistry - Class 12 NEET-2022-Ch-395 Solution: Find the emf of the cell in which the following reaction takes…

To determine the electromotive force (EMF) of the given cell reaction at 298 K: Reaction: Ni(s) + 2Ag + (0.001 M) _ Ni 2+ (0.001 M) + 2Ag(s) Given data: Standard cell potential, E� cell = 10.5 V Value at 298 K, 2.303 R T F = 0.059 Using the Nernst equation to find the cell potential, E cell : E c e l l = E c e l l o _ 0.059 n log _ [ Ni 2 + ] [ Ag + ] 2 $E_{cell} = E_{cell}^o - \frac{0.059}{n} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}$ E ce ll _ = E ce ll o _ _ n 0.059 _ lo g [ Ag + ] 2 [ Ni 2 + ] _ Where: n = number of moles of electrons transferred = 2 [ Ni 2 + ] = 0.001 M $[\text{Ni}^{2+}] = 0.001\, \text{M}$ [ Ni 2 + ] = 0.001 M [ Ag + ] = 0.001 M $[\text{Ag}^+] = 0.001\, \text{M}$ [ Ag + ] = 0.001 M Substituting the values: E c e l l = 10.5 _ 0.059 2 log _ ( 10 _ 3 ) ( 10 _ 3 ) 2 $E_{cell} = 10.5 - \frac{0.059}{2} \log \frac{(10^{-3})}{(10^{-3})^2}$ E ce ll _ = 10.5 _ 2 0.059 _ lo g ( 1 0 _ 3 ) 2 ( 1 0 _ 3 ) _ = 10.5 _ 0.0295 2 log _ ( 10 3 ) $= 10.5 - \frac{0.0295}{2} \log (10^3)$ = 10.5 _ 2 0.0295 _ lo g ( 1 0 3 ) = 10.5 _ 0.0295 _ 3 $= 10.5 - 0.0295 \times 3$ = 10.5 _ 0.0295 _ 3 = 10.5 _ 0.0885 $= 10.5 - 0.0885$ = 10.5 _ 0.0885 = 10.4115 V $= 10.4115\, \text{V}$ = 10.4115 V Therefore, the EMF of the cell is 10.4115 V.