Question 13 Marks
$E_{\text {cell }}^0$ for the given redox reaction is 2.71 V
$Mg _{( s )}+ Cu _{(0.01 M )}^{2+} \longrightarrow \stackrel{2++ Cu _{( s )}}{ Mg _{(0.001 M )}^{2+}}$
Calculate $E_{\text {cell }}$ for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and (ii) greater than 2.71 V .
$Mg _{( s )}+ Cu _{(0.01 M )}^{2+} \longrightarrow \stackrel{2++ Cu _{( s )}}{ Mg _{(0.001 M )}^{2+}}$
Calculate $E_{\text {cell }}$ for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and (ii) greater than 2.71 V .
Answer
View full question & answer→$E_{cell} = E_{cell}^{0} - \frac{0.059}{2} log \frac{[Mg^{2+}]}{[Cu^{2+}]}$
$E_{cell} = 2.71 - \frac{0.059}{2} log \frac{[0.001]}{[0.01]}$
$E_{cell} = 2.71 - (-0.0295) = 2.74 V$
(i) When external opposite applied voltage is less than 2.71 , it is less than $E _{\text {cell }}^0$, therefore, the electrons will flow from the anode to the cathode, and current will flow from cathode (copper electrode) to anode (magnesium electrode).
(ii) When external opposite applied potential is greater than 2.71, it is greater than $E _{\text {cell }}^0$, therefore, the reaction will be reversed, and the current will flow from anode to cathode.
$E_{cell} = 2.71 - \frac{0.059}{2} log \frac{[0.001]}{[0.01]}$
$E_{cell} = 2.71 - (-0.0295) = 2.74 V$
(i) When external opposite applied voltage is less than 2.71 , it is less than $E _{\text {cell }}^0$, therefore, the electrons will flow from the anode to the cathode, and current will flow from cathode (copper electrode) to anode (magnesium electrode).
(ii) When external opposite applied potential is greater than 2.71, it is greater than $E _{\text {cell }}^0$, therefore, the reaction will be reversed, and the current will flow from anode to cathode.