Question

If the following reactions are used to make a galvanic cell, which species will be reduced and which species will be oxidized?
$F_2+2e^{-}\to 2F^{-}\left(aq\right);E=+2.87V$
$C{a}^{+}+2e^{-}\to Ca\left(s\right);E=-2.76V$

• A. $F^{-}$ will be oxidized and $C{a}^{2+}$ will be reduced
• B. $C{a}^{2+}$ 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 oxidized 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 $C{a}^{+}$ is negative, $Ca\left(s\right)$ will be oxidised.

$F_2\left(g\right)+2Ca\left(s\right)\to 2C{a}^{+}\left(aq\right)+2F^{-}\left(aq\right)$