write the cell reaction and calculate the standard potential of the ce

1.	write the cell reaction and calculate the standard potential of the cell :zn/zn2+(1M)//cd2+(1M)/cd. Given Ezn/zn2+=0.763V,Ecd,cd2+=0.403V
Answer» ANSWER: STANDARD Potential of the cell is 0.36 V Explanation: $\rule{300}{1.5}$ Given information is: $\bullet\qquad\sf Zn\;\|\;Zn^{+2}\quad\parallel\quad Cd^{+2}\;\|\;Cd$ The two reactions are: $\sf 1)\quad Zn\longrightarrow Zn^{+2} + 2\;e^{-}\\\\\\ \sf 2) \quad Cd^{+2} + 2\;e^{-}\longrightarrow Cd$ And as we KNOW that losing of electrons is termed as Oxidation, and gain of electrons is termed as reduction. We know, At ANODE: Oxidation takes place. At Cathode: Reduction takes place. Therefore, Reaction at anode:- $\sf Zn\longrightarrow Zn^{+2} + 2\;e^{-}$ Reaction at cathode:- $\sf Cd^{+2} + 2\;e^{-}\longrightarrow Cd$ $\\$ $\begin{array}{c c} \sf{At\;Anode:}&\sf {Zn\longrightarrow Zn^{+2} + 2\;e^{-}}\\\\\sf{At\;Cathode:}&\sf{Cd^{+2} + 2\;e^{-}\longrightarrow Cd}\\\\\cline{1-2}\\\sf{Cell\;Reaction:}&\sf{Zn+Cd^{+2}\longrightarrow Zn^{+2}+Cd}\\\\\cline{1-2}\end{array}$ $\\$ So, the cell reaction is $\sf Zn+Cd^{+2}\longrightarrow Zn^{+2}+Cd$ $\rule{300}{1.5}$ $\rule{300}{1.5}$ Now we need to find standard potential of the cell i.e. Emf of cell. Therefore, we know, $\bigstar\;\underline{\boxed{\sf EMF_{(Cell)}=E^{\circ}_{(Cathode)}-E^{\circ}_{(Anode)}}}$ Substituting the values, $\longrightarrow\sf EMF_{(Cell)}= -0.403-(-0.763)\\\\\\\longrightarrow\sf EMF_{(Cell)}=-0.403+0.763\\\\\\\longrightarrow\sf EMF_{(Cell)}=0.36\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf EMF_{(Cell)}=0.36\;V}}}}$ ∴ Standard Potential of the cell is 0.36 V. $\\$ Note:- I have taken standard reduction potential but in the QUESTION standard oxidation potential is given. Therefore, you need to change it before substituting. $\rule{300}{1.5}$

write the cell reaction and calculate the standard potential of the cell :zn/zn2+(1M)//cd2+(1M)/cd. Given Ezn/zn2+=0.763V,Ecd,cd2+=0.403V

Answer»

ANSWER:

STANDARD Potential of the cell is 0.36 V

Explanation:

$\rule{300}{1.5}$

Given information is:

$\bullet\qquad\sf Zn\;|\;Zn^{+2}\quad\parallel\quad Cd^{+2}\;|\;Cd$

The two reactions are:

$\sf 1)\quad Zn\longrightarrow Zn^{+2} + 2\;e^{-}\\\\\\ \sf 2) \quad Cd^{+2} + 2\;e^{-}\longrightarrow Cd$

And as we KNOW that losing of electrons is termed as Oxidation, and gain of electrons is termed as reduction.

We know,

At ANODE: Oxidation takes place.
At Cathode: Reduction takes place.

Therefore,

Reaction at anode:- $\sf Zn\longrightarrow Zn^{+2} + 2\;e^{-}$

Reaction at cathode:- $\sf Cd^{+2} + 2\;e^{-}\longrightarrow Cd$

$\\$

$\begin{array}{c c} \sf{At\;Anode:}&\sf {Zn\longrightarrow Zn^{+2} + 2\;e^{-}}\\\\\sf{At\;Cathode:}&\sf{Cd^{+2} + 2\;e^{-}\longrightarrow Cd}\\\\\cline{1-2}\\\sf{Cell\;Reaction:}&\sf{Zn+Cd^{+2}\longrightarrow Zn^{+2}+Cd}\\\\\cline{1-2}\end{array}$

$\\$

So, the cell reaction is $\sf Zn+Cd^{+2}\longrightarrow Zn^{+2}+Cd$

$\rule{300}{1.5}$

Now we need to find standard potential of the cell i.e. Emf of cell.

Therefore, we know,

$\bigstar\;\underline{\boxed{\sf EMF_{(Cell)}=E^{\circ}_{(Cathode)}-E^{\circ}_{(Anode)}}}$

Substituting the values,

$\longrightarrow\sf EMF_{(Cell)}= -0.403-(-0.763)\\\\\\\longrightarrow\sf EMF_{(Cell)}=-0.403+0.763\\\\\\\longrightarrow\sf EMF_{(Cell)}=0.36\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf EMF_{(Cell)}=0.36\;V}}}}$

∴ Standard Potential of the cell is 0.36 V.

$\\$

Note:- I have taken standard reduction potential but in the QUESTION standard oxidation potential is given. Therefore, you need to change it before substituting.

$\rule{300}{1.5}$