JEE MAIN - Chemistry (2019 - 10th January Evening Slot - No. 11)
In the cell
Pt$$\left| {\left( s \right)} \right|$$H2(g, 1 bar)$$\left| {HCl\left( {aq} \right)} \right|$$AgCl$$\left| {\left( s \right)} \right|$$Ag(s)|Pt(s)
the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is :
$$\left\{ {} \right.$$Given, $${{2.303RT} \over F} = 0.06V$$ at $$\left. {298} \right\}$$
Pt$$\left| {\left( s \right)} \right|$$H2(g, 1 bar)$$\left| {HCl\left( {aq} \right)} \right|$$AgCl$$\left| {\left( s \right)} \right|$$Ag(s)|Pt(s)
the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is :
$$\left\{ {} \right.$$Given, $${{2.303RT} \over F} = 0.06V$$ at $$\left. {298} \right\}$$
0.94 V
0.40 V
0.76 V
0.20 V
Explanation
Anode : H2(g) $$ \to $$ 2H+(aq) + 2e-
Cathode : AgCl(s) + e- $$ \to $$ Ag(s) + Cl-(aq)
---------------------------------------------------------------
H2(g) + 2AgCl(s) $$ \to $$ 2Ag(s) + 2H+(aq) + 2Cl-(aq)
From Nernst equation we know,
Ecell = E0cell - $${{0.06} \over n}\log Q$$
Here,
Ecell = E0cell - $${{0.06} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}{{\left[ {C{l^ - }} \right]}^2}} \over {{P_{{H_2}}}}}$$
$$ \Rightarrow $$ 0.92 = E0AgCl/AgCl - - $$0.03\log {{{{\left[ {{{10}^{ - 6}}} \right]}^2}{{\left[ {{{10}^{ - 6}}} \right]}^2}} \over 1}$$
$$ \Rightarrow $$ E0AgCl/AgCl = 0.92 - 0.72 = 0.2 V
Cathode : AgCl(s) + e- $$ \to $$ Ag(s) + Cl-(aq)
---------------------------------------------------------------
H2(g) + 2AgCl(s) $$ \to $$ 2Ag(s) + 2H+(aq) + 2Cl-(aq)
From Nernst equation we know,
Ecell = E0cell - $${{0.06} \over n}\log Q$$
Here,
Ecell = E0cell - $${{0.06} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}{{\left[ {C{l^ - }} \right]}^2}} \over {{P_{{H_2}}}}}$$
$$ \Rightarrow $$ 0.92 = E0AgCl/AgCl - - $$0.03\log {{{{\left[ {{{10}^{ - 6}}} \right]}^2}{{\left[ {{{10}^{ - 6}}} \right]}^2}} \over 1}$$
$$ \Rightarrow $$ E0AgCl/AgCl = 0.92 - 0.72 = 0.2 V
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