The standard cell potential ΔE° is 1.100V for the cell Zn(s)/Zn²⁺//Cu²⁺/Cu(s) if [Zn²⁺] = 0.01M and [Cu²⁺] = 1.0M, what is the ΔE or EMF?

The cell notation $\text{Zn(s)}/\text{Zn}^{2+}//\text{Cu}^{2+}/\text{Cu(s)}$ indicates the following half-reactions:
Anode (oxidation): $\text{Zn(s)} \to \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-$
Cathode (reduction): $\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \to \text{Cu(s)}$
Overall reaction: $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \to \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$
From the half-reactions, the number of electrons transferred, $n$, is 2.

The Nernst equation at $25^\circ\text{C}$ is:

Where $\Delta E^\circ$ is the standard cell potential, $n$ is the number of electrons transferred, and $Q$ is the reaction quotient.

For the overall reaction $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \to \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$, the reaction quotient is:

Given concentrations are $[\text{Zn}^{2+}] = 0.01 \, M$ and $[\text{Cu}^{2+}] = 1.0 \, M$.

Given $\Delta E^\circ = 1.100 \, \text{V}$, $n=2$, and $Q=0.01$.

Rounding to three decimal places, $\Delta E = 1.159 \, \text{V}$.

Comparing this to the given options:

(a) $1.159 \, \text{V}$

(b) $0.923 \, \text{V}$

(c) $1.111 \, \text{V}$

(d) $1.030 \, \text{V}$

(e) $0.991 \, \text{V}$

The calculated value matches option (a).

The final answer is $\boxed{\text{(a) } 1.159\text{V}}$.