(a) Discuss the electrochemical theory of corroison
(b) For the reaction
`Ni(s)+2Ag^(+)(1M) to Ni^(2+)(1M)+2Ag(s)`
Which species gets reduced ?
1 Answers
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which of the following will increase the voltage of the cell with following cell reaction? `Sn(s)+2Ag^(+)+(aq)rarrSn^(+2)(aq)+2Ag(s)`
Correct Answer - C `E_("cell") = E_("cell")^(@) - (RT)/(nF) "ln" ([Sn^(2+)])/([Ag^(+)]^(2))` .
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In the reaction , `Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2Ag(s)`, the reduction half cell reaction is
Correct Answer - C Copper undergo oxdiation while silver `(Ag^(+))` undergo reduction.
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Calculate equilibrium constant for the reaction at `25^(@)C` `Cu(s)+2Ag^(+)(aq) hArr Cu^(2+)(aq)+2Ag(s)` `E^(@)` value of the cell is 0.46`" V "`.
`E_(cell)`=0.46 V, n=2 `log K_(c)=(nE_(cell)^(@))/(0.0591)=(2xx0.46)/(0.0591)=15.6` `K_(c)="Antilog " 15.6=3.98xx10^(15)`.
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Determine the values of equilibrium constant `(K_(c))` and `DeltaG^(@)` for the reaction `Ni(s)+2Ag^(+)(aq)rarrNi^(2+)(aq)+2Ag(s), E^(@)=1.05 V.` (`"G
(i) Calculate of `DeltaG^(@)` for the reaction `Ni(s)+2Ag^(+)(aq)rarr Ni^(2+)(aq)+2Ag(s)` `DeltaG^(@)=-nFE_(cell)^(@)` `=(-2)xx(96500" C mol"^(-1))xx(1.05 V)=-202650 "CV mol"^(-1)` =-202650 J `mol^(-1)`=-202.65 kJ `mol^(-1)` (ii) Calculate of equilibrium constant `(K_(c))` for the reaction...
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Depict the galvanic in which the reaction `:` `Zn(s)+2Ag^(+)(aq) rarr Zn^(2+)(aq)+2Ag(s)` takes place. Further show `:` ` a.` Which of the electrode i
The galvanic cell for the reaction is `Zn(s)|Zn^(2+)(aq)|| 2Ag^(+)(aq) | 2Ag(s)` (i) Zinc electrode (anode) is negatively charged. (ii) From silver to zinc in external circuit (iii) Anode :`" "...
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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)^(@)=
(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"...
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Calculate the following cell reaction cell at 298 K. `2Ag^(+)+Cd to 2Ag+Cd^(2+)` `E^(@)`for `Ag^(+)//Ag` and `Cd^(2+)//Cd` are 0.80 V and -0.40 V resp
Correct Answer - 1.18824V; Yes (i) The cell representation is : `Cd(s)+2Ag^(+)(aq) to Cd^(2+)(aq)+2Ag` (ii) `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cd^(2+)])/([Ag^(+)]^(2))` `E_(cell)=(1.2" V ")-((0.0591" V "))/(2)"log"(0.1)/(0.2xx0.2)` `=1.2V-(0.02955" V")xx0.3979` `=1.2-0.01176=1.18824" V "` (iii) Yes, Since `E_(cell)` is...
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Depict the galvanic cell in which the reaction `Zn(s) +2Ag^(+)(aq) to Zn^(2+)(aq)+2Ag(s)` takes place. Further indicate what are the carriers of curre
`Zn^(2+)` and `Ag^(+)` ions are carriers of current inside the cell. Electrons are the carriers of current outside the cell.
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For the reaction : `Ni(s)+2Ag^(+)(1 M) to Ni^(2+)(1 M) +2Ag(s)` which species gets reduced ?
`Ag^(+)` ions get reduced.
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For the cell reaction, `Mg(s)+2Ag^(+) (aq.) hArr Mg^(2+) (aq.)+2Ag(s)` `E_(cell)^(@)` is +3.17 V at 298 K. The value of `E_(cell), Delta G^(@)` and Q
Correct Answer - D `E^(@) = + 3.17V, n = 2` `Q = ([Mg^(2+)])/([Ag^(+)]^(2)) = (0.02)/([0.001]^(2)) = 20000` `Delta G^(@) = -nFE^(@)` `= - 2 xx 96500 xx 3.17` `= -611.8...
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