ElectrochemistryHard

Question

Two electrochemical cells are assembled in which the following reactions occur:

V2+ + VO2+ + 2H+ → 2V3+ + H2O; E°Cell = 0.616 V

V3+ + Ag+ + H2O → VO2+ + Ag(s) + 2H+; E°Cell = 0.439 V

If $E_{Ag^{+}|Ag}^{o}$= 0.799 V, what is$E_{V^{3 +}|V^{2 +}}^{o}$?

Options

A.−0.256 V
B.+0.256 V
C.+1.854 V
D.−1.854 V

Solution

$E_{V^{2 +}|V^{3 +}}^{o} = 1 \times 0.616 + 1 \times 0.439 - 1 \times 0.799 = 0.256\text{ V}$

$\therefore E_{V^{3 +}|V^{2 +}}^{o} = - 0.256\text{ V}$

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