Engineering Chemistry S Datta

ELECTROCHEMICAL CELLS 253
Highlights:
• If after calculation of E cell
it comes negative, half cells are to be reversed to get
positive value of E cell
. Here if Ag-electrode is made positive electrode (one half
cell) and Zn-electrode is made (negative electrode) (the other half cell) then E cell
would become positive.
• If we cannot make right choice during construction of a cell, E cell
would come out
negative, but the table of ∈° values and the above procedure will help to calculate
E cell
of any reversible cell.
Example 4. Calculate the e.m.f. of a Daniel cell at 25°C, when the concentrations of
ZnSO 4
and CuSO 4
are 0.001 M and 0.1 M, respectively. The standard potential of cell is 1.1 volts.
° °
Sol. We can write E cell
= ( ∈+
− ∈−
) + RT [Cu ]
nF
ln +
+2
[Zn ]
= E° Cell
+ 0.0592 0.1
log
2 0.001
[Q E° cell
= ∈° +
– ∈° –
]
= 1.1 + 0.0296 log 100
= 1.1 + 0.0296 × 2
= 1.1592 V.
Example 5. Calculate the e.m.f. of the cell:
Ni | Ni +2 1(M) || Pb +2 1(M) | Pb
at 25°C. Write down its cell reaction. Given ∈° Ni
= – 0.24 V and ∈° Pb
= – 0.13 V at 25°C.
Sol. E° cell
= ∈° right
– ∈° left
= (– 0.13) – (– 0.24)
= 0.11 V.
Cell reaction:
At anode: Ni – 2e Ni +2 (Ox)
At cathode: Pb +2 + 2e Pb (Red)
Ni + Pb +2 Ni +2 + Pb
Example 6. Sketch the electrochemical cell and write down the cell reaction, if Zn and
Ag electrodes are dipped in respective solutions. Also find out the standard emf of the cell, given
that ∈° + 2 = 0.76 V and ∈° + = 0.8 V at 25°C.
Zn/Zn Ag/Ag
Sol. ∈° Ox
= ∈° +
Zn/Zn 2 = 0.76 V i.e., ∈° Red =∈ Zn
+ 2 = – 0.76 V
/Zn
∈° Ox
= ∈° +
Ag/Ag = – 0.8 V i.e., ∈° Red = ∈° +
Ag / Ag
= 0.8 V.
So here S.R.P of Ag-electrode is higher than that of Zn-electrode. So, Ag-electrode will
form right hand half cell i.e., +ve electrode.
Cu = Zn/Zn +2 (a = 1)/Ag + (a = 1)/Ag +
According to ‘2R’ convention reduction will take place at right hand electrode (cathode).
Cell reaction ⇒
At cathode: 2Ag + + 2e 2Ag (Reduction)
At anode: Zn – 2e Zn +2 (Oxidation)
2Ag + + Zn +2 2Ag + Zn +2
2