1. Tardigrade
  2. Question
  3. Chemistry
  4. 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

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Q. 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

Electrochemistry

Solution:

$K _{ sp }$ for $AgCl =10^{-9} \times(0.1)=10^{-10}$
$\therefore E_{ Cl ^{-} / AgCl / Ag }^{0}=+0.80+\frac{0.059}{1} \log 10^{-10}=0.21\, V$