Question

Construct a cell using standard hydrogen electrode and zinc electrode. Write its cell reaction and cell representation. Calculate cell potential of a cell with 0.01 M Zn²⁺ ions. Standard potential of a cell is + 0.76 V.

Answer 1

Cell Reaction:

1. At the anode (zinc electrode), oxidation occurs:
   \[\ce{Zn_{(s)} -> Zn^2+_{(aq)} + 2e^-}\]
2. At the cathode (SHE), reduction occurs:
   \[\ce{2H^+_{(aq)} + 2 e^- -> H_2_{(g)}}\]
3. Overall Cell Reaction:
   \[\ce{Zn_{(s)} + 2H^+_{(aq)} -> Zn^2+_{(aq)} + H2_{(g)}}\]

Cell representation:

\[\ce{Zn_{(s)} | Zn^2+_{(aq)} || H^+_{(aq)} | H_2_{(g)} | Pt_{(s)}}\]

Given:

[Zn²⁺] = 0.01 M

E°_Zn = +0.76V

To find:

E_Z = ?

Solution:

\[\ce{Zn^2+_{(aq)}(0.01M) + 2e^- -> Zn_{(s)}}\]

`E_{Zn} = E°_{Zn} - 0.0592/n log_10  1/[[Zn^(2+)]]`

= `0.76 - (0.0592)/n log_10  1/0.01`

= 0.76 - 0.0592

= 0.7008

∴ E_zn = 0.7008