Question

(Equation 1) $F_2+2e^{-}\to 2F^{-}\left(aq\right)$
$E^o=+2.87V$
(Equation 2) ${Ca}^{2+}+2e^{-}\to Ca\left(s\right)$
$E^o=-2.76V$
If we combine the above half reactions in a galvanic cell, which species will be oxidized and which one will reduced ?

• A. $F^{-}$ will be oxidized and ${Ca}^{2+}$ will be reduced
• B. ${Ca}^{2+}$ will be oxidized and $F_2$ will be reduced
• C. $Ca\left(s\right)$ will be oxidized and $F_2$ will be reduced
• D. $F_2$ will be reduced and $Ca\left(s\right)$ will be reduced

Answer 1

The correct option is C $Ca\left(s\right)$ will be oxidized and $F_2$ will be reduced

As the reduction potential of $F_2$ is positive, $F_2$ will be reduced and as the reduction potential of $C{a}^{2+}$ is negative, $Ca\left(s\right)$ will be oxidized and $C{a}^{2+}$ will be formed after oxidation of $Ca\left(s\right)$.