A cell is prepared by dipping a copper rod in 1 M `CuSO_(4)` solution and a nickel rod in 1 M `NiSO_(4)` solution. The standard reduction potentials of copper electrode and nickel electrode are 0.34 volt and -0.25 volt respectively.
(a) What will be the cell reaction?
(b) What will be the stadnard EMF of the cell?
(c) Which electrode will be positive?
(d) How will the cell be represented?

Explanation

The two half cel reductio equation are :
`2Fe^(2+)(aq) e^(-) rarr Fe(s),E^(@)=-0.45V`
`Cr^(3+)(aq)+3 ^(-) rarr Cr(s),E^(@)=-0.75V`
Since `Cr^(3+)//Cr` electrode has lower electrode potential therfore it acts as the anode while `Fe^(2+)//Fe` electrode with higher electrode potential acts as the cathode
to equalise the number of electrons multiply Eq (i) by 3 and Eq (ii) by 2 but do not multiply their `E^(@)` values thus
To obtain equation for the cell reaction subtract Eq (iv) form Eq (iii) we have
`2 Cr (s) +3Fe^(2+)(aq) rarr 2cr^(3+) (aq)+3 Fe(s) ,E_(cell)^(@) =-0.45 -(-0.75V)=+0.30 V`
Thus the EmF of the cell =-+0.30 V