A) 0.38 volt
B) 1.28 volt
C) 0.32 volt
D) 0.66 volt
Correct Answer: C
Solution :
\[0.421\,={{E}^{\text{o}}}-\frac{0.059}{3}\log \,\frac{0.001}{{{(0.01)}^{3}}}\] \[{{E}^{\text{o}}}=0.421\,+\frac{0.059}{3}\log ({{10}^{3}})\] \[{{E}^{\text{o}}}\,=0.480\,=0.8\,-{{E}^{\text{o}}}{{M}^{+3}}/M\] \[{{E}^{\text{o}}}_{{{M}^{+3}}/M}=0.32\]You need to login to perform this action.
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