Question

Two electrochemical cells are assembled in which following reactions occur.

$V^{2+}+V{O}^{2+}+2H^{+}\to 2V^{3+}+H_{2}O;E_{cell}^o=0.616V$
$V^{3+}+Ag+2H_{2}O\to V{O}_2+4H^{+}+Ag\left(s\right);E_{cell}^o=0.439V$

$E^o$ for half reaction, $V^{3+}+e\to V^{2+}$ is :
Given that $E_{A{g}^{+}/Ag}=0.799V.$

• A. -0.256 V
• B. -0.216 V
• C. -0.346 V
• D. None of these

Answer 1

Answer: (A)

-0.256 V

Given,
$V^{2+}+V{O}^{2+}+2H^{+}\to 2V^{3+}+H_{2}O;E_{Icell}^o=0.616V$ ..... (i)

$V^{3+}+A{g}^{+}+2H_{2}O\to V{O}_2+4H^{+}+Ag\left(s\right);E_{IIcell}^o=0.439V$ ...... (ii)

By equation (i) $E_{Icell}^o=E_{O{P}_{V^{2+}/V^{3+}}}^o+E_{R{P}_{V^{4+}/V^{3+}}}^o$ ....... (iii)

By equation (ii) $E_{IIcell}^o=E_{O{P}_{V^{3+}/V^{4+}}}^o+E_{R{P}_{A{g}^{+}/Ag}}^o$ ...... (iv)

On adding Equation (iii) and (iv)

$(\because E_{O{P}_{V^{3+}/V^{4+}}}^o=-E_{R{P}_{V^{4+}/V^{3+}}}^o)$

$E_{Icell}^o+E_{IIcell}^o=E_{O{P}_{V^{2+}/V^{3+}}}^o+E_{R{P}_{A{g}^{+}/Ag}}^o$

$\therefore 0.616+0.439=E_{O{P}_{V^{2+}/V^{3+}}}^o+0.799$

$\therefore E_{O{P}_{V^{2+}/V^{3+}}}^o=+0.256V$

$E_{R{P}_{V^{3+}/V^{2+}}}^o=-0.256V$