InterviewSolution
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InterviewSolution
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The following chemical reaction occurring in an electrochemical cell : `Mg(s)+2Ag^(+)(0.0001 M) to Mg^(2+)(0.10 M)+2Ag(s)` Given `E_(Mg^(2+)//Mg)^(@)=-2.36" V", E_(Ag^(+)//Ag)^(@)=+0.81" V"` For the cell, calculate/write : (i) `E^(@)` value for the electrode `2Ag^(+)//2Ag` (ii) Standard cell potential `(E^(@))` (iii) Cell potential (E) (iv) Give the symbolic representation of the above cell (v) Will the above cell reaction be spontaneous ? |
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Answer» (i) `E^(@)` value of the electrode , `2Ag^(+)//2Ag` will remain the same i.e., 0.81 V. It does not change. (ii) `" " E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=0.81-(-2.36)=3.17" V"` (iii) `" " E_(cell)=E_(cell)^(@)=((0.0591" V"))/(2)"log"([Mg^(2+)(aq)])/([Ag^(+)(aq)]^(2))` `=(3.17" V")-((0.0591" V"))/(2)"log"((0.1))/((0.0001)^(2)` `=(.17" V")-(0.2068" V")=2.9632" V" ` (iv) Cell notation :` Mg(s)//Mg^(2+)(0.10" M") || 2Ag^(+)(0.0001" M")//2Ag(s)` (v) Since Mg is placed below Ag in the activity series, it is a stronger reducing agent. Therefore, the cell reaction is spontaneous. |
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