JEE MAIN - Chemistry (2020 - 2nd September Evening Slot - No. 1)
For the disproportionation reaction
2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10–1.
Given :
($$E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$$
$$E_{C{u^ + }/Cu}^0 = 0.52V$$
$${{RT} \over F} = 0.025$$)
2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10–1.
Given :
($$E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$$
$$E_{C{u^ + }/Cu}^0 = 0.52V$$
$${{RT} \over F} = 0.025$$)
Answer
144
Explanation
$$E_{cell}^0$$ = $$E_{C{u^ + }/Cu}^0$$ - $$E_{C{u^{2 + }}/C{u^ + }}^0$$
= 0.52 – 0.16
= 0.36 V
At equilibrium, Ecell = 0
$$E_{cell}^0$$ = $${{RT} \over {nF}}$$ln K
$$ \Rightarrow $$ ln K = $${{E_{cell}^0 \times nF} \over {RT}}$$
= $${{0.36 \times 1} \over {0.025}}$$ = 14.4 = 144 $$ \times $$ 10-1
= 0.52 – 0.16
= 0.36 V
At equilibrium, Ecell = 0
$$E_{cell}^0$$ = $${{RT} \over {nF}}$$ln K
$$ \Rightarrow $$ ln K = $${{E_{cell}^0 \times nF} \over {RT}}$$
= $${{0.36 \times 1} \over {0.025}}$$ = 14.4 = 144 $$ \times $$ 10-1
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