Question

What is the standard cell notation of the Galvanic cell made with silver and nickel?

Answer 1

Hint :Start with the anode components (site of oxidation) at the left side and then the cathode components (site of reduction) are listed to the right. Components of the right side and left side are separated by \[||\] which represents the salt bridge.
 Anode \[||\] Cathode

Complete Step By Step Answer:
Reduction potential for half cell of \[A{g^ + }\]and \[N{i^{2 + }}\]are as follows:
\[A{g^ + }(aq) + 1{e^ - } \to Ag(s)\] \[{E^0} = 0.80\;V\] \[\langle 1\rangle \]
\[N{i^{2 + }}(aq) + 2{e^ - } \to Ni(s)\] \[{E^0} = - 0.23\;V\] \[\langle 2\rangle \]
The value of standard electrode potential of a cell is given by
\[E_{cell}^0 = E_{cathode}^0 - E_{anode}^0\]
In the galvanic cell, the reaction is spontaneous that is value of \[\vartriangle {G^0}\] and for a spontaneous reaction \[E_{cell}^0\] must be a positive quantity, since \[\vartriangle {G^0} = - nF{E^0}\].
Now, we have to manipulate the two equations in such a way that the electrons lost at anode should be equal to the electrons gained at cathode, we get
\[\{ A{g^ + }(aq) + 1{e^ - } \to Ag(s)\} \times 2\] \[{E^0} = 0.80\;V\]
 \[Ni(s) \to N{i^{2 + }}(aq) + 2{e^ - }\] \[{E^0} = 0.23\;V\]
\[\overline {\underline {2A{g^ + }(aq) + Ni(s) \to 2Ag(s) + N{i^{2 + }}(aq)} } \] \[\overline {\underline {E_{cell}^0 = 1.03V} } \]
        Hence, standard cell notation of galvanic cell made with silver and nickel is
\[Ni(s)|N{i^{2 + }}(aq)\;||A{g^ + }(aq)|Ag(s)\]
The single vertical lines indicate the boundary (phase difference) between solid \[Ni\]and \[N{i^{2 + }}\] ions in the aqueous solution of the first compartment and between solid \[Ag\] and \[A{g^ + }\] ions present in the aqueous solution of the second compartments.

Additional Information:
The value of standard electrode potential of a cell can be calculated in this way also,
\[E_{cell}^0 = E_{cathode}^0 - E_{anode}^0\]
For given cell, value of standard electrode potential is
\[E_{cell}^0 = E_{A{g^ + }}^0 - E_{N{i^{2 + }}}^0\]
\[E_{cell}^0 = 0.80V - ( - 0.23)V\]
\[E_{cell}^0 = 1.03V\]

Note :
Values of \[{E^0}\] will be provided in the question, no need to mug up. Also, electrochemical series should be known to know which species will undergo oxidation and which species will undergo reduction. Out of the two species given, the one with a higher value of reduction potential will undergo reduction and will be present at the cathode.