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A cell is prepared by dipping a chromium rod in 1 MCr_(2)(sO_(4))_(3) solution and an iron rod in 1 M FeSO_(4) solution The standard reduction potentials of chromium and iron electrodes ae -0.75 V and -0.45 V respecitvely (a) what will be the cell reactoition ? (b) what will be the standard EMF of hte cell ? (C ) which electrode will act as anode ? (d) which electrode will act as cathods ?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The two half cel reductio equation are : <br/> `2Fe^(2+)(aq) e^(-) rarr Fe(s),E^(@)=-0.45V` <br/> `Cr^(3+)(aq)+3 ^(-) rarr Cr(s),E^(@)=-0.75V` <br/> Since `Cr^(3+)//Cr` electrode has lower electrode <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> therfore it acts as the anode while `Fe^(2+)//Fe` electrode with higher electrode potential acts as the cathode <br/> to equalise the number of electrons <a href="https://interviewquestions.tuteehub.com/tag/multiply-1106275" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLY">MULTIPLY</a> Eq (i) by 3 and Eq (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) by 2 but do not multiply their `E^(@)` values <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a> <br/> To obtain equation for the cell reaction subtract Eq (iv) form Eq (iii) we have <br/> `2 Cr (s) +3Fe^(2+)(aq) rarr 2cr^(3+) (aq)+3 Fe(s) ,E_(cell)^(@) =-0.45 -(-0.75V)=+0.30 V` <br/> Thus the EmF of the cell =-+0.30 V</body></html>


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