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 ?

A cell is prepared by dipping a chromium rod in 1 MCr_(2)(sO_(4))_(3)

1.	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» SOLUTION :The two half cel reductio equation are : `2Fe^(2+)(aq) e^(-) rarr Fe(s),E^(@)=-0.45V` `Cr^(3+)(aq)+3 ^(-) rarr Cr(s),E^(@)=-0.75V` Since `Cr^(3+)//Cr` electrode has lower electrode POTENTIAL therfore it acts as the anode while `Fe^(2+)//Fe` electrode with higher electrode potential acts as the cathode to equalise the number of electrons MULTIPLY Eq (i) by 3 and Eq (II) by 2 but do not multiply their `E^(@)` values THUS To obtain equation for the cell reaction subtract Eq (iv) form Eq (iii) we have `2 Cr (s) +3Fe^(2+)(aq) rarr 2cr^(3+) (aq)+3 Fe(s) ,E_(cell)^(@) =-0.45 -(-0.75V)=+0.30 V` Thus the EmF of the cell =-+0.30 V