Equilibrium constant -K-c- the given cell reaction is 10 - calculate E-circ-cell- - A-s- - B-2-aq- rightleftharpoons A-2- -aq- - B-s-

Question

Equilibrium constant  -K-c-  the given cell reaction is  10 - calculate  E-circ-cell-  -  A-s- - B-2-aq- rightleftharpoons A-2- -aq- - B-s-

Toppr Admin · Accepted Answer

- A -s-160- - B-2- -aq- - -160-rightleftharpoons- - A-2-aq- - B-s- -160-160- -160- -160- Here - - - n - 2 - K-c - 10 -160-160- -160- -160- - E-o-cell- - -dfrac-0-059-n- -times log -K-c -160-160- -160- -160- -160- E-o-cell- - -dfrac-0-059-n- -times log-10 - 0-0295-V -160-

Equilibrium constant $$ (K_c) $$ for the given cell reaction is $$ 10 $$. calculate $$ E^{\circ}_{cell} $$ .
$$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+} (aq) + B(s) $$

$$ A (s) + B^{2+} (aq) \, \, \rightleftharpoons\, \, A^{2+}(aq) + B(s) $$
Here , , $$ n = 2 , K_c = 10 $$
$$ E^{o}_{cell} = \dfrac{0.059}{n} \times log \,K_c $$
$$ E^{o}_{cell} = \dfrac{0.059}{n} \times log\,10 = 0.0295\,V $$

Equilibrium constant $$ (K_c) $$ for the given cell reaction is $$ 10 $$. calculate $$ E^{\circ}_{cell} $$ . $$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+} (aq) + B(s) $$

$$ A (s) + B^{2+} (aq) \, \, \rightleftharpoons\, \, A^{2+}(aq) + B(s) $$ Here , , $$ n = 2 , K_c = 10 $$ $$ E^{o}_{cell} = \dfrac{0.059}{n} \times log \,K_c $$ $$ E^{o}_{cell} = \dfrac{0.059}{n} \times log\,10 = 0.0295\,V $$

Equilibrium constant $$ (K_c) $$ for the given cell reaction is $$ 10 $$. calculate $$ E^{\circ}_{cell} $$ .
$$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+} (aq) + B(s) $$

$$ A (s) + B^{2+} (aq) \, \, \rightleftharpoons\, \, A^{2+}(aq) + B(s) $$
Here , , $$ n = 2 , K_c = 10 $$
$$ E^{o}_{cell} = \dfrac{0.059}{n} \times log \,K_c $$
$$ E^{o}_{cell} = \dfrac{0.059}{n} \times log\,10 = 0.0295\,V $$