Solution:

Given, the electrode potential of silver (Ag) is 0.80 V and that of copper (Cu) is 0.34 V.

The standard reduction potential of a reaction is the tendency of a species to undergo reduction, compared to the standard hydrogen electrode (SHE). The reduction potential of SHE is zero.

The reduction half-reaction for silver is:

Ag+ + e- → Ag (E°red = 0.80 V)

The oxidation half-reaction for copper is:

Cu → Cu2+ + 2e- (E°red = -0.34 V)

The overall cell reaction can be obtained by adding the two half-reactions:

Ag+ + Cu → Ag + Cu2+

The overall cell potential (Ecell) can be calculated using the Nernst equation:

Ecell = E°cell - (0.0592/n) log(Q)

where E°cell is the standard cell potential, n is the number of electrons transferred in the cell reaction, and Q is the reaction quotient.

In this case, n = 2, since two electrons are transferred in the cell reaction.

The reaction quotient can be expressed as:

Q = [Ag+][Cu2+]/[Ag][Cu]

Since the cell is at standard conditions, the concentrations of Ag+ and Cu2+ are 1 M, and the concentrations of Ag and Cu are 1 M. Therefore, Q = 1.

Substituting the values in the Nernst equation, we get:

Ecell = E°cell - (0.0592/2) log(1)

Ecell = E°cell

Therefore, the cell potential (Ecell) is equal to the standard cell potential (E°cell).

The standard cell potential can be obtained by subtracting the reduction potential of the anode (Cu) from the reduction potential of the cathode (Ag):

E°cell = E°cathode - E°anode

E°cell = 0.80 V - (-0.34 V)

E°cell = 1.14 V

Therefore, the e.m.f. of the galvanic cell is 1.14 V.

Hence, option (C) is the correct answer.

Electrode potential of Ag is greater than Cu
Ag is cathode and Cu is anode
emf = E(cathode) - E(anode) = 0.8 - 0.34 = 0.46V