If the concentration of Mg2+ is 0.01 M, the electrode potential of Mg2+/Mg electrode will be (Given that Eo Mg2+ Mg=-2.36V ):

Question

If the concentration of Mg2+ is 0.01 M, the electrode potential of Mg2+/Mg electrode will be (Given that Eo Mg2+ Mg=-2.36V ): (A)  +2.80  (B)  -2.80  (C)  +2.41  (D)  -2.41

Tardigrade Admin · Accepted Answer

EoMg2+,Mg=EoMg2+,Mg =(0.0591/2) log ([Mg+2]/[Mg]) =-2.36+(0.0591/2) log (0.01/1) =-2.36+(0.0591/2) log 0.01 =-2.36-0.0591 =-2.4191

Q. If the concentration of $M g^{2 +}$ is 0.01 M, the electrode potential of $M g^{2 +} / M g$ electrode will be (Given that $E^{o} M g^{2 +}$ $M g = - 2.36 V$ ):

$+ 2.80$

$- 2.80$

$+ 2.41$

$- 2.41$

$E^{o}_{M g^{2 +}},_{M g} = E^{o}_{M g^{2 +}, M g}$ $= \frac{0.0591}{2} lo g \frac{[ M g ^{+ 2} ]}{[ M g ]}$ $= - 2.36 + \frac{0.0591}{2} lo g \frac{0.01}{1}$ $= - 2.36 + \frac{0.0591}{2} lo g 0.01$ $= - 2.36 - 0.0591$ $= - 2.4191$

Q. If the concentration of Mg2+ is 0.01 M, the electrode potential of Mg2+/Mg electrode will be (Given that EoMg2+ Mg=−2.36V ):

+2.80

−2.80

+2.41

−2.41