For the galvanic cell,
Zn(s) + Cu$$^{2+}$$(0.02M) $$\rightarrow$$ Zn$$^{2+}$$(0.04M) + Cu(s)
E$$_{cell}$$ = _________ $$\times 10^{-2}$$ V (Nearest integer)
[Use: E$$^0$$ Cu/Cu$$^{2+}$$ = $$-0.34$$ V, E$$_{Zn/Zn^{2+}}$$ = +0.76 V, $$\frac{2.303RT}{F}$$ = 0.059 V]

We have the galvanic cell

$$\text{Zn}(s)+\text{Cu}^{2+}(0.02\,\text M)\;\longrightarrow\;\text{Zn}^{2+}(0.04\,\text M)+\text{Cu}(s).$$

For any galvanic cell the standard e.m.f. is obtained from the standard reduction potentials of the two half-cells by the relation

$$E^{\circ}_{\text{cell}} \;=\; E^{\circ}_{\text{cathode}} \;-\; E^{\circ}_{\text{anode}}.$$

The data supplied are

$$E^{\circ}_{\text{Cu}/\text{Cu}^{2+}}=-0.34\;{\rm V},\qquad E^{\circ}_{\text{Zn}/\text{Zn}^{2+}}=+0.76\;{\rm V}.$$

These are oxidation potentials. To convert them into reduction potentials we merely change the sign:

$$E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = -(-0.34)=+0.34\;{\rm V},$$ $$E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -(+0.76)= -0.76\;{\rm V}.$$

The copper couple acts as the cathode (reduction) and the zinc couple as the anode (oxidation). Hence

$$E^{\circ}_{\text{cell}} = (+0.34)\;-\;(-0.76)=1.10\;{\rm V}.$$

Next we use the Nernst equation. The general form for a cell with $$n$$ electrons transferred is

$$E = E^{\circ}_{\text{cell}} - \frac{0.059}{n}\,\log Q,$$

where $$Q$$ is the reaction quotient. In the present reaction

$$\text{Zn}(s)+\text{Cu}^{2+}(aq)\;\longrightarrow\;\text{Zn}^{2+}(aq)+\text{Cu}(s)$$

the solids have unit activity, so

$$Q=\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} =\frac{0.04}{0.02}=2.$$

Two electrons are exchanged, so $$n=2$$. Substituting every value into the Nernst equation gives

$$E = 1.10 - \frac{0.059}{2}\,\log(2).$$

We now evaluate the logarithmic term. Using $$\log(2)=0.3010$$ (base-10),

$$\frac{0.059}{2}\,\log(2)=0.0295\times 0.3010 = 0.0089\;{\rm V}\;(\text{to four significant figures}).$$

Therefore

$$E = 1.10 - 0.0089 = 1.0911\;{\rm V}.$$

Written as a multiple of $$10^{-2}$$ volts,

$$1.0911\;{\rm V}=109.11\times 10^{-2}\;{\rm V}\;\approx\;109\times 10^{-2}\;{\rm V}\;(\text{nearest integer}).$$

Hence, the correct answer is Option 109.