Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95℅ dissociated at this dilution at 298K.calculate the electrode Potential given that E¢ ( Zn2+|Zn) = - 0.76 V.

Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95℅ disso

1.	Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95℅ dissociated at this dilution at 298K.calculate the electrode Potential given that E¢ ( Zn2+\|Zn) = - 0.76 V.
Answer» <html><body><p><strong>Answer:</strong></p><p>The electrode reaction is :</p><p>Zn2+(<a href="https://interviewquestions.tuteehub.com/tag/aq-883254" style="font-weight:bold;" target="_blank" title="Click to know more about AQ">AQ</a>)+2e−→Zn(s)</p><p>According to Nernst <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> :</p><p>E=E∘−0.0591nlog1[Mn+(aq)]</p><p>E=E∘−0.05912log1[Zn2+]</p><p>Since 0.1 M ZnSO4 solution is <a href="https://interviewquestions.tuteehub.com/tag/95-342378" style="font-weight:bold;" target="_blank" title="Click to know more about 95">95</a>% <a href="https://interviewquestions.tuteehub.com/tag/dissociated-7675483" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOCIATED">DISSOCIATED</a>, [Zn2+]=0.95×0.1=0.095M.</p><p>E=−0.76−0.05912log1(0.095)</p><p>=−0.76+0.02955 <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> 0.095=−0.76−0.02955×1.0223</p><p>=−0.76−0.0302=−0.7902V.</p><p><strong>Explanation:</strong></p><p></p></body></html>