Question

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

• A. $ +2.80 $
• B. $ -2.80 $
• C. $ +2.41 $
• D. $ -2.41 $

Answer 1

Answer: (D)

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