JEE MAIN - Chemistry (2020 - 8th January Evening Slot - No. 11)
For an electrochemical cell
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio $${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$$ when this cell attains equilibrium is _________.
(Given $$E_{S{n^{2 + }}|Sn}^0 = - 0.14V$$,
$$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$$, $${{2.303RT} \over F} = 0.06$$)
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio $${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$$ when this cell attains equilibrium is _________.
(Given $$E_{S{n^{2 + }}|Sn}^0 = - 0.14V$$,
$$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$$, $${{2.303RT} \over F} = 0.06$$)
Answer
2.13TO2.16
Explanation
Cell reaction is :
Sn(s) + Pb+2(aq) $$ \to $$ Sn+2(aq) + Pb(s)
Apply Nernst equation :
Ecell = $$E_{cell}^0$$ - $${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$ ....(1)
$$ \Rightarrow $$ $$E_{cell}^0$$ = -0.13 + 0.14 = 0.01 V
At equilibrium : Ecell = 0
Substituting in (1), we get
0 = 0.01 - $${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$
$$ \Rightarrow $$ $$\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$ = $${1 \over 3}$$
$$ \Rightarrow $$ $${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$$ = 2.15
Sn(s) + Pb+2(aq) $$ \to $$ Sn+2(aq) + Pb(s)
Apply Nernst equation :
Ecell = $$E_{cell}^0$$ - $${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$ ....(1)
$$ \Rightarrow $$ $$E_{cell}^0$$ = -0.13 + 0.14 = 0.01 V
At equilibrium : Ecell = 0
Substituting in (1), we get
0 = 0.01 - $${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$
$$ \Rightarrow $$ $$\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$ = $${1 \over 3}$$
$$ \Rightarrow $$ $${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$$ = 2.15
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