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  • A zinc rod is dipped in 0.1 M solution of   . The salt is 82 \% dissocia

A zinc rod is dipped in 0.1 M solution of  \mathrm{ZnSO}_4 . The salt is 82 \% dissociated at thisdilution at 298 K. Calculate electrode potential.
\left[\mathrm{E}_{\mathrm{zn}^{2+}}^{\mathrm{0}} / \mathrm{zn}=-0.76 \mathrm{~V}\right]

Option: 1

-0.64 V


Option: 2

-1.62 V


Option: 3

-0.72 V


Option: 4

-2.97 V


Answers (1)

best_answer

\mathrm{Zn}^{+2}+2 e^- \rightarrow \mathrm{Zn}
Using Nerst eq

\mathrm{ {\left[zn^{+2}\right]=\frac{82}{100} \times 0.1=0.082} }
\mathrm{ E _{Zn{ }^{+2} / Z n}=-0.76-0.0295 \log \frac{1}{0.082} }

\mathrm{ =-0.76-0.0295[\log 1000-\log 82] \\ }
\mathrm{=0.76-0.0295[3-1.91] \\ }
\mathrm{ =-0.72 ~Volt }.

Posted by

Divya Prakash Singh

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