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Question: The emf of the cell, \(Ni|N{i^{ + 2}}(1.0M)||A{g^ + }(1.0M)|Ag\) (\({E^ \circ }\) for \(N{i^{ + 2}}/...

The emf of the cell, Ni∣Ni+2(1.0M)∣∣Ag+(1.0M)∣AgNi|N{i^{ + 2}}(1.0M)||A{g^ + }(1.0M)|Ag (E∘{E^ \circ } for Ni+2/NiN{i^{ + 2}}/Ni = -0.25 volt, E∘{E^ \circ } for Ag+/AgA{g^ + }/Ag = 0.80 volt) is given by:
(A)-0.25 +0.80 =0.55 volt
(B)-0.25-(+0.80) = -1.05 volt
(C)+0.80 – (-0.25) = +1.05 volt
(D)-0.80 – (-0.25) = -0.55 volt

Explanation

Solution

The emf of a cell is the difference between the standard reduction potentials of cathode and anode. So, first write down the entire reaction and find out which one is cathode and anode.
Formula used:
The formula we will be using here is:
E∘cell=E∘cathode−E∘anode{E^ \circ }_{cell} = {E^ \circ }_{cathode} - {E^ \circ }_{anode}
Where E∘cell{E^ \circ }_{cell} = standard emf of cell;
E∘cathode{E^ \circ }_{cathode} =standard reduction potential of cathode;
E∘anode{E^ \circ }_{anode} = standard reduction potential of anode.

Complete-step- by- step answer:
The emf or the electromotive force of a cell is defined as the maximum potential difference between the two electrodes of the cell (that is between cathode and anode).
First of all let’s write the complete cell reaction for this cell:

                                   $Ni + 2A{g^ + } \to N{i^{ + 2}} + 2Ag$

Always remember that the element which undergoes oxidation forms the anode and it is written to the left while representing the cell and the element undergoing reduction forms the cathode and is written to the right while representing the cell.
So, from this and from the reaction we can tell that Ni is the anode (it undergoes oxidation) and Ag is the cathode (it undergoes reduction).
Their half cell reactions can be written as:

                                   Anode: $Ni \to N{i^{ + 2}} + 2{e^ - }$   
                                Cathode: $A{g^ + } + {e^ - } \to Ag$

It is given in the question the standard reduction potentials of both electrodes:
E∘anode{E^ \circ }_{anode} = E∘{E^ \circ } of Ni+2/NiN{i^{ + 2}}/Ni = -0.25 volt
E∘cathode{E^ \circ }_{cathode} = E∘{E^ \circ } of Ag+/AgA{g^ + }/Ag = 0.80 volt

We know that the value given is standard reduction potential because it says Ni+/NiN{i^ + }/Ni and Ag+/AgA{g^ + }/Ag which shows us that the value is for the reaction proceeding from Ni+N{i^ + } to Ni and Ag+A{g^ + } to Ag, both of which are reduction reactions.
And standard emf of a cell is given as:
E∘cell=E∘cathode−E∘anode{E^ \circ }_{cell} = {E^ \circ }_{cathode} - {E^ \circ }_{anode}
= 0.80 – (-0.25)
= 0.80 + 0.25
= +1.05 volt

So, the correct option is: (C) +0.80 – (-0.25) = +1.05 volt.

Note: While doing such questions always build the complete reaction correctly and find out the anode and the cathode. Also while putting values in the emf equation always check that the values of cathode and anode should both should be either oxidation potential or both should be reduction potential.