Question

What is ${\text{G}}^{\text{o}}$ at 298 K for the reaction, ${\text{2VO}}_2^{+}(aq)+4H^{+}(aq)+Cd(s)\to 2V{O}^{2+}(aq)$ $+2H_{2}O(l)+C{d}^{2+}(aq)$

• A. $-271kJ$
• B. $1.403J$
• C. $-135kJ$
• D. $-115kJ$

Answer 1

Answer: (A)

$2e^{-}+2V{O}_2^{+}+4H^{+}\to 2V{O}^{2+}+2H_{2}O;$ $\ast \Delta E^o=1.0V$ ?(i) $Cd\to C{d}^{2+}2e^{-};$ ${}^{\ast }\Delta E^o=+0.40V$ ?(ii) On adding Eq (i) and (ii), we get $2V{O}_2^{+}+4H^{+}+Cd\to 2V{O}^{2+}+2H_{2}O+C{d}^{2+};$ $E_{cell}^o=1.0+0.40=1.40V$ $\Delta G^o=-nF{E}_{cell}^o=-2\times 96500\times 1.4$ $=-270200J=-271kJ$ * Since in this $\text{Q}\text{.}{\text{E}}^{\text{o}}$ values are not given, it is incorrect. These values are taken from reference books.