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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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$E^o_{Zn^{+2}/Zn}=-0.76V$.So $E^o_{Zn/Zn^{+2}}=$?

$\begin{array}{1 1}(a)\;-0.076-\large\frac{0.0591}{2}\normalsize\log\large\frac{1}{Zn^{+2}}\\(b)\;-0.76\\(c)\;0.76\\(d)\;0\end{array}$

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For any electrode
Reduction potential= - oxidation potential
So $E^o_{Zn/Zn^{+2}}=-E^o_{Zn^{+2}/Zn}=0.76V$
Hence (c) is the correct answer.
answered Dec 10, 2013 by sreemathi.v

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