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The emf of a galvanic cell constituted with the electrodes `Zn^(2+)//Zn(E^(@)=-0.76" V")` and `Fe^(2+)//Fe(E^(@)=-0.41" V")` is :A. `-0.35" V"`B. `+1.17" V"`C. `+0.35" V"`D. `-1.17" V"` |
Answer» Correct Answer - C (c ) `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)` `=(-0.41)-(-0.76)=0.35" V"` |
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