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

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

Explanation

Solution

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 Ag+A{g^ + }and Ni2+N{i^{2 + }}are as follows:
Ag+(aq)+1eAg(s)A{g^ + }(aq) + 1{e^ - } \to Ag(s) E0=0.80  V{E^0} = 0.80\;V 1\langle 1\rangle
Ni2+(aq)+2eNi(s)N{i^{2 + }}(aq) + 2{e^ - } \to Ni(s) E0=0.23  V{E^0} = - 0.23\;V 2\langle 2\rangle
The value of standard electrode potential of a cell is given by
Ecell0=Ecathode0Eanode0E_{cell}^0 = E_{cathode}^0 - E_{anode}^0
In the galvanic cell, the reaction is spontaneous that is value of G0\vartriangle {G^0} and for a spontaneous reaction Ecell0E_{cell}^0 must be a positive quantity, since G0=nFE0\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
Ag+(aq)+1eAg(s)×2\\{ A{g^ + }(aq) + 1{e^ - } \to Ag(s)\\} \times 2 E0=0.80  V{E^0} = 0.80\;V
Ni(s)Ni2+(aq)+2eNi(s) \to N{i^{2 + }}(aq) + 2{e^ - } E0=0.23  V{E^0} = 0.23\;V
2Ag+(aq)+Ni(s)2Ag(s)+Ni2+(aq)\overline {\underline {2A{g^ + }(aq) + Ni(s) \to 2Ag(s) + N{i^{2 + }}(aq)} } Ecell0=1.03V\overline {\underline {E_{cell}^0 = 1.03V} }
Hence, standard cell notation of galvanic cell made with silver and nickel is
Ni(s)Ni2+(aq)  Ag+(aq)Ag(s)Ni(s)|N{i^{2 + }}(aq)\;||A{g^ + }(aq)|Ag(s)
The single vertical lines indicate the boundary (phase difference) between solid NiNiand Ni2+N{i^{2 + }} ions in the aqueous solution of the first compartment and between solid AgAg and Ag+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,
Ecell0=Ecathode0Eanode0E_{cell}^0 = E_{cathode}^0 - E_{anode}^0
For given cell, value of standard electrode potential is
Ecell0=EAg+0ENi2+0E_{cell}^0 = E_{A{g^ + }}^0 - E_{N{i^{2 + }}}^0
Ecell0=0.80V(0.23)VE_{cell}^0 = 0.80V - ( - 0.23)V
Ecell0=1.03VE_{cell}^0 = 1.03V

Note :
Values of E0{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.