JEE MAIN - Chemistry (2020 - 6th September Evening Slot - No. 16)
For the given cell :
Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s)
change in Gibbs energy ($$\Delta $$G) is negative, if :
Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s)
change in Gibbs energy ($$\Delta $$G) is negative, if :
C2 = $$\sqrt 2 $$C1
C2 = $${{{C_1}} \over {\sqrt 2 }}$$
C1 = 2C2
C1 = C2
Explanation
Given $$\Delta $$G < 0
$$ \therefore $$ -nFEcell < 0
$$ \Rightarrow $$ Ecell > 0
We know, Ecell = $$E_{cell}^0$$ - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$
= 0 - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$
$$ \therefore $$ - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$ > 0
$$ \Rightarrow $$ $$\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$ < 0
$$ \Rightarrow $$ C1 < C2
By checking option, we can see
C2 = $$\sqrt 2 $$C1 satisfy the condition C1 < C2.
$$ \therefore $$ -nFEcell < 0
$$ \Rightarrow $$ Ecell > 0
We know, Ecell = $$E_{cell}^0$$ - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$
= 0 - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$
$$ \therefore $$ - $${{RT} \over {2F}}\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$ > 0
$$ \Rightarrow $$ $$\ln \left( {{{{C_1}} \over {{C_2}}}} \right)$$ < 0
$$ \Rightarrow $$ C1 < C2
By checking option, we can see
C2 = $$\sqrt 2 $$C1 satisfy the condition C1 < C2.
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