We are asked to calculate for the **E _{cell} for the reaction**.

**Cu(s) | Cu ^{2+}(0.017 M) || Ag(s), (Ag^{+} = 0.18M)**

Recall that the** Nernst Equation** relates the concentrations of compounds and cell potential.

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{05916}\mathbf{}\mathbf{V}}{\mathbf{n}}\mathbf{\right)}{\mathbf{}}{\mathbf{log}}{\mathbf{}}{\mathbf{Q}}}$

E_{cell} = cell potential under non-standard conditions

E°_{cell} = standard cell potential

n = mole e^{-} transferred

Q = reaction quotient = products/reactants

We're going to calculate for the E_{cell} using the following steps:

*Step 1:** **Identify the anode and the cathode in the reaction and write the overall reaction*** Step 2: **Calculate E

What is E of the following cell reaction at 25°C? E°_{cell} = 0.460 V.

Cu(s) | Cu^{2+}(0.017 M) || Ag(s), (Ag^{+} = 0.18 M)

A) 0.468V

B) 0.282 V

C) 0.460 V

D) 0.490 V

E) 0.479V

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