JEE MAIN - Chemistry (2019 - 11th January Evening Slot - No. 11)
Given the equilibrium constant:
KC of the reaction :
Cu(s) + 2Ag+ (aq) $$ \to $$ Cu2+ (aq) + 2Ag(s) is
10 $$ \times $$ 1015, calculate the E$$_{cell}^0$$ of this reaciton at 298 K
[2.303 $${{RT} \over F}$$ at 298 K = 0.059V]
KC of the reaction :
Cu(s) + 2Ag+ (aq) $$ \to $$ Cu2+ (aq) + 2Ag(s) is
10 $$ \times $$ 1015, calculate the E$$_{cell}^0$$ of this reaciton at 298 K
[2.303 $${{RT} \over F}$$ at 298 K = 0.059V]
0.4736 mV
0.04736 V
0.4736 V
0.04736 mV
Explanation
We know,
$$\Delta $$Go = -RTln(KC) ....(1)
Also $$\Delta $$Go = -nF$$E_{cell}^o$$ ....(2)
$$ \therefore $$ -nF$$E_{cell}^o$$ = -RTln(KC)
$$ \Rightarrow $$ $$E_{cell}^o$$ = $${{RT} \over {nF}}\ln \left( {{K_C}} \right)$$
= $$2.303{{RT} \over {nF}}\log \left( {{K_C}} \right)$$
= $${{0.059} \over 2}\log \left( {10 \times {{10}^{15}}} \right)$$
( n = no of electron transferred = 2 )
= 0.059 $$ \times $$ $${{16} \over 2}$$
= 0.059 $$ \times $$ 8
= 0.472 V
$$\Delta $$Go = -RTln(KC) ....(1)
Also $$\Delta $$Go = -nF$$E_{cell}^o$$ ....(2)
$$ \therefore $$ -nF$$E_{cell}^o$$ = -RTln(KC)
$$ \Rightarrow $$ $$E_{cell}^o$$ = $${{RT} \over {nF}}\ln \left( {{K_C}} \right)$$
= $$2.303{{RT} \over {nF}}\log \left( {{K_C}} \right)$$
= $${{0.059} \over 2}\log \left( {10 \times {{10}^{15}}} \right)$$
( n = no of electron transferred = 2 )
= 0.059 $$ \times $$ $${{16} \over 2}$$
= 0.059 $$ \times $$ 8
= 0.472 V
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