At 298 K take E^°Ag^+/Ag=+0.80 V…

At 298 K take E^°Ag^+/Ag=+0.80 V, Ksp(AgCl)=1.8×10^-10, (2.303RT)/(F)=0.0592 V, R=8.314 J mol^-1K^-1 and log 1.8=0.255.

One or more options correct

298\ \mathrm{K}

take

E^\circ_{\mathrm{Ag^+/Ag}}=+0.80\ \mathrm{V}

K_{sp}(\mathrm{AgCl})=1.8\times10^{-10}

\dfrac{2.303RT}{F}=0.0592\ \mathrm{V}

R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}

and

\log 1.8=0.255

. The correct statement(s) is/are

A. $E^\circ_{\mathrm{AgCl/Ag,Cl^-}}\approx+0.22\ \mathrm{V}$
B. A silver wire dipped in a saturated solution of AgCl in pure water has an electrode potential of about $+0.22\ \mathrm{V}$
C. $\Delta G^\circ$ for $\mathrm{AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)}$ is about $-55.6\ \mathrm{kJ\,mol^{-1}}$
D. When $[\mathrm{Cl^-}]$ is decreased from $1.0\ \mathrm{M}$ to $0.10\ \mathrm{M}$ , the potential of the $\mathrm{Ag|AgCl|Cl^-}$ electrode falls to about $0.16\ \mathrm{V}$

Show answer and solution

Answer: A

The silver-silver chloride couple is obtained by combining

\mathrm{Ag^+ + e^-\to Ag}

with the solubility equilibrium:

E^\circ_{\mathrm{AgCl/Ag}}=E^\circ_{\mathrm{Ag^+/Ag}}+0.0592\log K_{sp}=0.80+0.0592(-9.745)=+0.223

V, so A is correct. B: in saturated AgCl without added chloride,

[\mathrm{Ag^+}]=\sqrt{K_{sp}}=1.34\times10^{-5}

M, so

E=0.80+0.0592\log(1.34\times10^{-5})=+0.51

V, not

0.22

V; B is incorrect. C:

\Delta G^\circ

for dissolution is

-RT\ln K_{sp}=+55.6

kJ mol

^{-1}

(positive, as the salt is sparingly soluble), so the stated negative sign is wrong. D: the electrode reaction

\mathrm{AgCl+e^-\to Ag+Cl^-}

has

E=E^\circ-0.0592\log[\mathrm{Cl^-}]

; lowering

[\mathrm{Cl^-}]

tenfold raises

E

0.0592

V to about

0.28

V, not

0.16

V. Only A is correct.

Original question written in the JEE Advanced 2021 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake