You visited us 1 times! Enjoying our articles? Unlock Full Access!

EMF

Knowing that ...

Question

Knowing that $K_{s p}$ for AgCl is $1.0 \times 10^{- 10}$ , Calculate E for a silver/silver chloride electrode immersed in $1.00 M K C l a t 25^{\circ} C$ . $E_{{A g}^{+} | A g}^{-} = 0.799 V$ .

Open in App

Solution

$A g + C l^{-} \to A g C l + e^{-}$
$E = E^{0} -$ $\frac{0.0591}{1}$ $log \frac{1}{[{C l}^{-}]}$
$= 0.799 - 0.0591 log (\frac{1}{105})$ $= 0.5035 V .$

Suggest Corrections

Similar questions

A g_{2} S O_{4} (a q .)

0.711 V

K_{s p} (A g_{2} S O_{4})

E_{A g^{+} / A g}^{\circ} = 0.799 V)

A g_{2} S O_{4 (a q)} .

0.711 V

K s p (A g_{2} S O_{4})

E_{A g^{+} / A g}^{\circ} = 0.799 V]

0.1 M H C l

A g C l

- 0.208 V

E_{A g^{+} / A g}^{\circ} = - 0.799 V

K_{s p}

A g C l

0.10 M

N a C I

K_{s p}

A g C l = 1.20 \times 10^{- 10}

A g_{2} S O_{4} (a q)

K_{s p} (A g_{2} S O_{4})

x \times 10^{- 3}

E_{A g^{+} / A g}^{o} = 0.799 V

100 x

Join BYJU'S Learning Program