If at 298 K, the solubility of AgCl in 0.05 M BaCl 2 (completely dissociated) is found to be very nearly 10-9 M and E Ag / Ag +0=-0.80 V. Then the value of E Cl - / AgCl /Ag 0 (in volt) at the same temperature will be

Question

If at 298 K, the solubility of AgCl in 0.05 M BaCl 2 (completely dissociated) is found to be very nearly 10-9 M and E Ag / Ag +0=-0.80 V. Then the value of E Cl - / AgCl /Ag 0 (in volt) at the same temperature will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

K sp for AgCl =10-9 ×(0.1)=10-10 ∴ E Cl - / AgCl / Ag 0=+0.80+(0.059/1) log 10-10=0.21 V

Q. If at $298 K$ , the solubility of $A g Cl$ in $0.05 M$ $B a C l_{2}$ (completely dissociated) is found to be very nearly $1 0^{- 9} M$ and $E_{A g / A g^{+}}^{0} = - 0.80 V$ . Then the value of $E_{C l^{-} / A g Cl / A g}^{0}$ (in volt) at the same temperature will be

$K_{s p}$ for $A g Cl = 1 0^{- 9} \times (0.1) = 1 0^{- 10}$
$∴ E_{C l^{-} / A g Cl / A g}^{0} = + 0.80 + \frac{0.059}{1} lo g 1 0^{- 10} = 0.21 V$

Q. If at 298K, the solubility of AgCl in 0.05M BaCl2​ (completely dissociated) is found to be very nearly 10−9M and EAg/Ag+0​=−0.80V. Then the value of ECl−/AgCl/Ag0​ (in volt) at the same temperature will be

Ksp​ for AgCl=10−9×(0.1)=10−10 ∴ECl−/AgCl/Ag0​=+0.80+10.059​log10−10=0.21V

Q. If at $298 K$ , the solubility of $A g Cl$ in $0.05 M$ $B a C l_{2}$ (completely dissociated) is found to be very nearly $1 0^{- 9} M$ and $E_{A g / A g^{+}}^{0} = - 0.80 V$ . Then the value of $E_{C l^{-} / A g Cl / A g}^{0}$ (in volt) at the same temperature will be

$K_{s p}$ for $A g Cl = 1 0^{- 9} \times (0.1) = 1 0^{- 10}$
$∴ E_{C l^{-} / A g Cl / A g}^{0} = + 0.80 + \frac{0.059}{1} lo g 1 0^{- 10} = 0.21 V$