Question

Write redox couples involved in the reaction (A) to (D).

A.$Cu+Z{n}^{2+}\to C{u}^{2+}+Zn$

B.$Mg+F{e}^2\to M{g}^{2+}+Fe$

C. $B{r}_2+2C{l}^{-}\to C{l}_2+2B{r}^{-}$

D.$Fe+C{d}^{2+}\to Cd+F{e}^{2+}$

Answer 1

A.$Cu+Z{n}^{2+}\to C{u}^{2+}+Zn$

Here, $Cu$ is oxidized to $C{u}^{2+}$ and $Z{n}^{2+}$ is reduced to $Zn$.

Hence, the redox couples are $C{u}^{2+}/Cu$ and $Z{n}^{2+}/Zn$.

B.$Mg+F{e}^{2+}\to M{g}^{2+}+Fe$

Here $Mg$ is oxidized to $M{g}^{2+}$ and $F{e}^{2+}$

Hence, the redox couples are $M{g}^{2+}/Mg$ and $F{e}^{2+}/Fe$.

C.$B{r}_2+2C{l}^{-}\to C{l}_2+2B{r}^{-}$

Here $C{l}^{-}$ is oxidized to $C{l}_z$ and $B{r}_2$ is reduced to $B{r}^{-}$

Hence, the redox couples are $B{r}_2/B{r}^{-}$ and $C{l}_2/C{l}^{-}$

D.$Fe+C{d}^{2+}\to Cd+F{e}^{2+}$

Here, $Fe$ is oxidized to $F{e}^{2+}$ and $C{d}^{2+}$ is reduced to Cd.

Hence, the redox couples are $F{e}^{2+}/Fe$ and $C{d}^{2+}/Cd$.