Question

For the half reaction
$B\left(s\right)\to B^{2+}+2e^{-}$ $E_1^{\circ }=-0.44V$
and $B_2+\to B^{3+}+e^{-}$ $E_2^{\circ }=1.3V$
What is $E^{\circ }$ for the reaction,
$3e^{-}+B^{+3}\to B\left(s\right)$

• A. $0.86$ V
• B. $-0.14$ V
• C. $0.14$ V
• D. $-0.28$ V

Answer 1

Answer: (B)

$B\left(s\right)\to B^2+{2e}^{-}{E}_1^0=-0.44V$ $-i$

$B^{2+}\to B^{3+}+e^{-}{E}_1^0=1.3V$ $-ii$

$B^{3+}+{3e}^{-}\to B\left(s\right)E_3^0=?$ $-iii$

iii $=-(II+I)$

Or, $-3\times E_3^0\times F=-\{-1\times 1.3\times F+0.44\times 2\times F\}$

$\therefore E_3^0=-0.14V$