How would you determine the reduction potential of `Zn//Zn^(2+)(aq)`?

Explanation

(i) Consider the zinc electrode dipped in zinc sulphate solution using SHE

(ii) Step 1 :
The following galvanic cell is constructed using SHE
`Zn_((s))|Zn_((aq.1M))^(2+)||H_((aq.1M))^(+)|H_(2(g, 1 atm))|Pt_((s))`
(iii) Step 2:
The emf of the above galvanic cell is measured using a voltmeter. In this case the measured emf of the above galvanic cell is 0.76 V .
(iv) Calculation:
We know that,
`E_("cell")^(@) = (E_("ox")^@)_(Zn//Zn^(2+)) + (E_("red")^(@))_(SHE)`
`:. E_("cell")^(@) = 0.76 + 0V`
`= 0.76 V`
The oxidation potential corresponds to the below mentioned half cell reaction which takes place at the cathode.
`Zn to Zn^(2+) + 2e^(-)` (oxidation)
(v) The emf of the reverse reaction will give the reduction potential
`Zn^(2+) + 2e^(-) to Zn " " (E_("cell")^@)_(Zn^(2+)|Zn) = - 0.76V`