Question

If the half-reaction: Fe3+ + e- = Fe2+ were chosen as the standard reduction potential table reference instead of: 2H+ + 2e- = H2, what would the Eo value for Zn2+ reduction be?

Answer 1

Answer:

$E^{0}$ for $Zn^{2+}\mid Zn$ system would be -1.53 V

Explanation:

Determination of standard reduction potential of any system is done by placing the system in cathode and a reference half cell in anode and then evaluate the cell potential. The cell potential is the standard reduction potential of the system.

So $E_{cell}^{0}=E_{system}^{0}-E_{reference}^{0}$

As $E_{H^{+}\mid H_{2}}^{0}$ is equal to 0 therefore cell potential is equal to reduction potential of any system by taking hydrogen electrode as a reference.

$E_{Zn^{2+}\mid Zn}^{0}$ equal to -0.76 V with respect to hydrogen

$E_{Fe^{3+}\mid Fe^{2+}}^{0}$ equal to 0.77 V with respect to hydrogen

Therefore standard reduction potential of $Zn^{2+}\mid Zn$ system when $Fe^{3+}\mid Fe^{2+}$ system is taken as reference is-

$E_{cell}^{0}=E_{Zn^{2+}\mid Zn}^{0}-E_{Fe^{3+}\mid Fe^{2+}}^{0}$

                          = -0.76 V - 0.77 V

                          = -1.53 V