A) \[+2.80\]
B) \[-2.80\]
C) \[+2.41\]
D) \[-2.41\]
Correct Answer: D
Solution :
\[{{E}^{o}}_{M{{g}^{2+}}}{{,}_{Mg}}={{E}^{o}}_{M{{g}^{2+}},Mg}\] \[=\frac{0.0591}{2}\log \frac{[M{{g}^{+2}}]}{[Mg]}\] \[=-2.36+\frac{0.0591}{2}\log \frac{0.01}{1}\] \[=-2.36+\frac{0.0591}{2}\log 0.01\] \[=-2.36-0.0591\] \[=-2.4191\]You need to login to perform this action.
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