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A cell is prepared by dipping a chromium rod is 1 M Cr_2(SO_4)_3 solution and an iron and in 1M FeSO_4 solutions. The standard potentials of chromium and iron electrodes are -0.75 V and 0.45 respectively. (a) What will be the cell reaction? (b) What will the standard EMF of the cell? (c) Which electrode will act an anode? (d) WHich electrode will act as cathode?

Answer» <html><body><p></p>Solution :The two half cell <a href="https://interviewquestions.tuteehub.com/tag/reduction-621019" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCTION">REDUCTION</a> equations are: <br/> `Fe^(2+)(aq)+2e^(-) to Fe(s),E^@=-0.45V`....(i) <br/> `<a href="https://interviewquestions.tuteehub.com/tag/cr-427229" style="font-weight:bold;" target="_blank" title="Click to know more about CR">CR</a>^(3+)(aq)+3e^(-)toCr(s),E^@=-0.75 V`.....(ii) <br/>Since `Cr^(3+)//Cr` electrode has a power reduction <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> it acts as the anode, while `Fe^(2+)//Fe` electrode with a higher electrode potential acts as the cathode. <br/> To equilize the number of electrons, multiply Eq (i) by 3 and Eq. (ii) by 2. But do not multiply their `E^@` values . Thus, <br/> `3Fe^(2+)(aq)+6e^(-)toFe(s),E^@=-0.45V`.....(<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) <br/> `2Cr^(3r)(aq)+6e^(-)to2Cr(s),E^@=-0.75V`.....(iv)<br/> To obtain equations for the cell reactions, <a href="https://interviewquestions.tuteehub.com/tag/subtract-1231765" style="font-weight:bold;" target="_blank" title="Click to know more about SUBTRACT">SUBTRACT</a> Eq. (iv) from Eq. (iii) , we have, <br/> `2Cr(s)+3Fe^(2+)(aq)to2Cr^(3+)(aq)+3Fe(s)` <br/> `E_(cell)^o=-0.45-(-0.075V)=+0.30V` Thus, the EMF of the cell =+0.30 V</body></html>


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