Consider the figure below and answer the...

Consider the figure below and answer the following questions :
(i) Cell 'A' has `E_(cell)=2" V "` and Cell B has `E_(cell) =1.1" V "` which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell ?
(ii) If cell 'A' has `E_(cell)=0.5" V "` and cell 'B' has `E_(cell) =1.1" V "` then what will be the reactions at anode and cathode ?

Text Solution

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(i) Since the cell (B) has low e.m.f. , it will act as electrolytic cell. In the cell, the electrode reaction that takes place is :
`Zn^(2+)(aq)+2e^(-) to Zn(s)" "("At cathode")`
`Cuoverset((aq)) rarr Cu^(2+)(aq)+2e^(-)`
(ii) The data shows that the cell 'B' will act as electrochemical cell since its `E_(cell)`. (1.1V) is mre than that of cell 'A' (0.5V).The electrode reactions which take place are :
`Zn(s) to Zn^(2+)(aq)+2e^(-)" " ("At anode")`
`Cu^(2+)(aq)+2e^(-) to Cu(s)" " ("At cathode")`

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In which of the following cell(s): E_("cell")=E_("cell")^(@) ?

`Cu(s)|Cu^(2+)(0.01M)||Ag^(+)(0.1M)|Ag(s)`

`Pt(H_(2))|pH=1||Zn^(2+)(0.01M)||Zn(s)`

`Pt(H_(2))|pH=1||Zn^(2+)(1M)||Zn(s)`

`Pt(H_(2))|H^(+)=0.1M||Zn^(2+)(0.01)|||Zn(s)`

E_("cell")^(@)=1.1V for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?

1.1=`K_(c)`

`(2.303RT)/(2F)"log"K_(c)=1.1`

`"log" K_(c)=(2.2)/(0.059)`

`"log" K_(c)=1.1`

For a cell involving two electrons changes, E_(cell)^(@)=0.3" V" at 25^(@)C . The equilibrium constant for the reaction is :

`10^(10)`

`3xx10^(-2)`