Question

Consider the cell;
$V^{2+}+V{O}^{2+}+2H^{+}\to 2V^{3+}+H_{2}O;E^o=0.616V$

$V^{3+}+A{g}^{+}+H_{2}O\to V{O}^{2+}+2H^{+}+Ag;E^o=0.439V$
Potential of $E_{V^{3+}/V^{2+}}^o$ is :
(Given that $E_{A{g}^{+}/Ag}^o=0.799V$)

• A. $-0.256V$
• B. $0.256V$
• C. $1.055V$
• D. $0.177V$

Answer 1

Answer: (A)

$-0.256V$

The e.m.f of the cell for the second reaction is 0.439 V.

Hence,

$E_{cell}^0=E_{cathode}^0-E_{anode}^0=E_{A{g}^{+}|Ag}^0-E_{V{O}^{2+}|V^{3+}}^0$

$0.439V=0.799V-E_{V{O}^{2+}|V^{3+}}$

$E_{V{O}^{2+}|V^{3+}}=0.36V$

The e.m.f of the cell for the first reaction is 0.616 V.

Hence,

$E_{cell}^0=E_{cathode}^0-E_{anode}^0=E_{V{O}^{2+}|V^{3+}}^0-E_{V^{3+}|V^{2+}}^0$

$0.616V=0.36V-E_{V^{3+}|V^{2+}}^0$

$E_{V3+|V2+}^0=-0.256V.$

Hence, option A is the right answer.