Q.10. Standard reduction potential for Cu
^{2+}/Cu is +0.34 v. Calculate reduction potential at P
^{H} = 14
K
_{sp} of Cu(OH)
_{2} is 1.0 x 10
^{19}
Ans: For Cu(OH)
_{2}, K
_{sp} = [Cu
^{2+}][on
^{}]
^{2}
P
^{H} = 14 So, [H
^{+}) = 10
^{14}
K
_{w} = [H
^{+})[OH
^{})
[OH
^{}) =
= 1
So, [Cu
^{2+}] =
= 1.0 x 10
^{19} = 1 x 10
^{19}
E
_{cell} = E
^{0}_{cell} +
log
_{10}
= 0.34 +
log
_{10}[Cu
^{2+}]
= 0.34 +
log
_{10}[1 x 10
^{19}]
E
_{cell} =  0.2205 v at P
^{H} = 14
Medium Type
Q.1. In a fuel cell H
_{2} & O
_{2} react to produceelectricity. In process H
_{2} gas is oxidized at anode& O
_{2} at cathode. If 67.26 of H
_{2} atSTP reacts in 15 min. What is avg. current produced ? If entire current is used for electrodeposition of Cu from Cu
^{2+}, how many gm of Cu are deposited ?
Cathode: O
_{2} + 2H
_{2}O + 4e
^{} 4OH
^{}
Anode: H
_{2} + 2OH
^{} + 4e
^{} 2H
_{2}O + 2e
^{}
Ans: Moles of H
_{2} reacting =
= 3
Eq. of H
_{2} used = Moles x nfactor
= 3 x 2 = 6
i = 643.33 A
Also, Eq. of H
_{2} = Eq. of Cu formed
w
_{Cu} = 6 x
= 190.5 g
wt. of Cu deposited = 190.5 g
Q.2. 19 g fused SnCl
_{2} was electrolysed using inert electrodes. 0.119 g Sn was deposited at cathode. If nothing was given out during electrolysis, calculate ratio of wt. of SnCl
_{2} & SnCl
_{2} in fused state after electrolysis (at. et. Sn = 119).
Ans: electrolysis SnCl
_{2} yieldes:
Anode: 2Cl
^{} Cl
_{2} + 2e
^{}
Cathode: Sn
^{2+} + 2e
^{} Sn
Further Cl
_{2} formed at anode reacts with SnCl
_{2} to give SnCl
_{4}
SnCl
_{2} + Cl
_{2} SnCl
_{4}
Eq. of SnCl
_{2} lost during electrolysis = Eq. of Cl
_{2} formed during electrolysis = Eq. of Sn formed.
Eq. of Cl
_{2} formed =
= 2 x 10
^{3}
Eq. of SnCl
_{4} formed = 2 x 10
^{3} (
whole Cl
_{2} react to from SnCl
_{4})
Now, total loss in Eq. of SnCl
_{4} during complete course
= Eq. of SnCl
_{2} lost during electrolysis + Eq. of SnCl
_{2} lost during reaction with Cl
_{2}
= 2 x 10
^{3} + 2 x 10
^{3} = 4 x 10
^{3}
Initial eq. of SnCl
_{2} left in sol. = 2 x 10
^{1}  4 x 10
^{3} = 0.196
Eq. of SnCl
_{4} formed = 2 x 10
^{3} = 0.002
Q.3. Calculate emf of given cell rxn. and Pb(s) + Hg
_{2}So
_{4} PbSO
_{4}(s) + 2H
_{g}(l)
Given
= 0.126
=  0.789 v
K
_{SP} of PbSO
_{4} = 2.43 x 10
^{8} K
_{SP} of Hg
_{2}SO
_{4} = 1.46 x 10
^{6}.
Ans: Oxidation potential =  REduction Potential
Anode: Pb
Pb
^{2+} + 2e
^{2e}
Cathode: Hg
_{2}^{2+} + 2e
^{} HG
_{2}
E
_{cell} = E
^{0}_{cell} 
log
= (

) 
log
= 0.789  ( 0.126) 
log
K
_{SP} = [Pb
^{2+}][SO
_{4}^{2}] = [Pb
^{2+}]
^{2} [
[Pb
^{2+}] = [SO
_{4}^{2}]]
[Pb
^{2+}] =
Similarly,
E
_{cell} = 0.789 + 0.126 
log
= 0.941
Q.4. Two weak acid sol. HA
_{1} & HA
_{2} each with same conc. & having P
^{Ka} values 3 & 5 are placed in contact with Hydrogen electrode (1atm, 25
^{0}c) and are interconnected through salt bridge. Find emf of cell.
Ans: Cell is
Pt H
_{2}(1 atm)HA
_{2}HA
_{1}(H
_{2})(1 atm) P
_{t}
At L.H.S.,
= 0  0.059 (P
^{H})
_{2}
At R.H.S.,
=  0.059 (P
^{H})
_{1}
For acid HA
_{1},
HA
_{1} H
^{+} + A
_{1}^{}
[H
^{+}) = c
=
if
is small)
(P
^{H})
_{1} =
PK
_{a} 
log
_{10}c
Similarly,
(P
^{H})
_{2} =
PK
_{a} 
logc
E
_{cell} =  0.059(P
^{H})
_{1} + 0.059(P
^{H})
_{2}
= 0.059[
P

P
] =
[5  3]
= 0.059 v
Q.5. REduction potential diagram for Cu in acid solution is:
Calculate value of X.
Ans: Given, Cu
^{2+} + e
^{} Cu
^{+};
= 0.15 v ;
........................... (i)
Cu
^{+} + e
^{} Cu
^{+} ,
= 0.5 v ;
........................... (ii)
Cu
^{2+} + 2e
^{} Cu ;
= ? ;
........................... (iii)
=  n
F =  1 x 0.15 F =  0.15 F
= n
F =  1 x 0.5 x F =  0.5 F
Adding
+
=
 0.15 F + ( 0.5 F) =
=  0.65 F
 n
F =  0.65 F
=
= 0.325 volt.
X = 0.325 Volt