Question

Calculate the emf of a Cd-Cu cell in which Cd is in contact with 0.002 M CdSO4 and Cu in contact with 0.02 M CuSO4 solution. The standard emf of the cell is 0.74 V at 298 K.

Answer 1

The cell reaction is

$Cd +Cu^{+2} \to Cd^{+2}+Cu$

Number of electrons used$=2$

$E_{cell}=E_0-\frac{0.0591}{n} \space log_{10} \frac{[Cd^{+2}][Cu]}{[Cu^+][Cd]}$

$[Cd]=[Cu]=1$

$E_{cell}=0.74-\frac{0.0591}{2} \space log_{10} \frac{[Cd^{+2}]}{[Cu^{+2}]}$

$E_{cell}=0.74-\frac{0.0591}{2} \space log_{10} \frac{0.002}{0.02}=0.74+0.0295=0.7695 V$