Question

A zinc rod is placed in 0.1 MZnSO4 solution at 298 K. Write the electrode reaction and calculate the potential of the electrode. (EZ24/2 = -0.76V.)

Answer 1

Answer: The potential of electrode is -0.79 V

Explanation:

When zinc is dipped in zinc sulfate solution, the electrode formed is $Zn^{2+}(aq.)/Zn(s)$

Reduction reaction follows:  $Zn^{2+}(0.1M)+2e^-\rightarrow Zn(s);(E^o_{Zn^{2+}/Zn}=-0.76V)$

To calculate the potential of electrode, we use the equation given by Nernst equation:

$E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = -0.76 V

n = number of electrons exchanged = 2

$[Zn]=1M$    (concentration of pure solids are taken as 1)

$[Zn^{2+}]=0.1M$

Putting values in above equation, we get:

$E_{cell}=-0.76-\frac{0.059}{2}\times \log(\frac{1}{0.1})\\\\E_{cell}=-0.79V$

Hence, the potential of electrode is -0.79 V