Question

Find out the $E_{cell}^0$ from the given data
a.$Zn|Z{n^{2 + }}\parallel C{u^{2 + }}|Cu\,:\,\,E_{cell}^0 = 1.10\,V$
b.$Cu|C{u^{2 + }}\parallel A{g^ + }|Ag\,:\,\,E_{cell}^0 = 0.46\,V$
c.$Zn|Z{n^{2 + }}\parallel A{g^ + }|Ag\,:\,\,E_{cell}^0 = ?$
(Given $E_{C{u^{2 + }}|Cu}^0$ $ = 0.34\,V$ )
$(1)$ $ - 0.04\,V$
$(2)$ $ + 0.04\,V$
$(3)$ $ + 0.30\,V$
$(4)$ $1.56\,V$

Answer 1

Hint: The standard electrode potential of an electrochemical cell is represented by $E_{cell}^0$ .It is the potential of the cell at standard conditions - $1M$ concentration, $1\,bar$ pressure and ${25^ \circ }C$ . It can be calculated as the difference between standard reduction potentials of two half cells – cathode and anode.
Formula used:
$E_{cell}^0\, = \,E_{cathode}^0 - E_{anode}^0$

Complete answer: To find the $E_{cell}^0$ of the reaction (c), we must find the values of $E_{Z{n^{2 + }}|Zn}^0$ and $E_{A{g^ + }|Ag}^0$ and apply it in the above formula.
Let us start with the first cell reaction (a) $Zn|Z{n^{2 + }}\parallel C{u^{2 + }}|Cu\,:\,\,E_{cell}^0 = 1.10\,V$
$Zn\,\,\,\, \to \,\,\,\,Z{n^{2 + }} + 2e$ : Anode
$C{u^{2 + }} + 2e\,\,\, \to \,\,Cu$ : Cathode
$Zn + C{u^{2 + }} \to Z{n^{2 + }} + Cu$
Here, Zn electrode is getting oxidised to $Z{n^{2 + }}$ and $C{u^{2 + }}$ is getting reduced to Cu by donating and accepting two electrons respectively. Zn electrode is the anode and Cu electrode is the cathode.
 By using the formula,
\[ \Rightarrow \] $E_{cell}^0\, = \,E_{cathode}^0 - E_{anode}^0$
\[ \Rightarrow \] $E_{cell}^0$ $ = E_{C{u^{2 + }}|Cu}^0 - E_{Z{n^{2 + }}|Zn}^0$
It is given that $E_{C{u^{2 + }}|Cu}^0$ $ = 0.34\,V$
Therefore, $E_{Z{n^{2 + }}|Zn}^0$ $ = 0.34 - 1.10 = - 0.76\,V$
The half-cell reactions in (b) $Cu|C{u^{2 + }}\parallel A{g^ + }|Ag\,:\,\,E_{cell}^0 = 0.46\,V$ are:
$Cu\,\, \to \,\,C{u^{2 + }} + 2e$ : Anode
$A{g^ + } + 1e \to Ag$ : Cathode
To maintain electrical neutrality, we must multiply the values in silver cathode by two.
Therefore the cathodic reaction becomes:
$2A{g^ + } + 2e \to 2Ag$
And the overall cell reaction is: $Cu + 2A{g^ + } \to C{u^{2 + }} + 2Ag$
 Here, \[ \Rightarrow \] $E_{cell}^0$ $ = E_{A{g^ + }|Ag}^0 - E_{C{u^{2 + }}|Cu}^0$
$E_{A{g^ + }|Ag}^0$ $ = $ $0.46 + 0.34 = 0.80\,V$
Now we have to find the standard cell potential of the cell reaction: (c) $Zn|Z{n^{2 + }}\parallel A{g^ + }|Ag\,$.
The half-cell reactions are:
$Zn\,\, \to \,Z{n^{2 + }} + 2e$ : Anode
$2A{g^ + } + 2e\,\, \to \,\,2Ag$ : Cathode
$Zn + 2A{g^ + } \to Z{n^{2 + }} + 2Ag$
The cathode reaction is multiplied by two as in reaction (b) to maintain electrical neutrality.
By applying the formula:
$E_{cell}^0\, = \,E_{cathode}^0 - E_{anode}^0$
$E_{cell}^0 = E_{A{g^ + }|Ag}^0 - E_{Z{n^{2 + }}|Zn}^0$$$
$E_{cell}^0$$ = 0.80 - ( - 0.76) = 1.56\,V$
The right option is $(4)$ $1.56\,V$

Note:
The electrode in which oxidation takes place is known as an anode and the electrode in which reduction takes place is known as cathode.
The reduction potential of a given species will be the negative value of its oxidation potential.
The value of $E_{cell}^0$ becomes zero at equilibrium. At equilibrium, $E_{cathode}^0$ will be equal to $E_{anode}^0$ .